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# i need help:""<

hii,
i have cs assignment i tried to solve it but i still have many
errors , plzz help mee :"< it`s not cheating becuz i`ve tried & wrote
the prog. i just wanna you to show me my mistakes

#these are the operations
[a, b] + [c, d] = [a+c, b+d],
[a, b] - [c, d] = [a-d, b-c],
[a, b] * [c, d] = [min(ac, ad, bc, bd), max(ac, ad, bc, bd)],
1/[a, b] = [1/b, 1/a] only if 0 not in [a,b].

& the Question is to Write the class interval that has lower and upper
as data members, set, get, print, and constructors as member
functions. The class also has the four member functions add, subtract,
multiply, and divide.

Implement the class members, and write a driver that declares two
objects of the class interval and prints the results of the four
operations on the two declared objects.

Sample input / output:

Enter the lower and upper limits of the first interval: 2 8
[2, 8]
Enter the lower and upper limits of the second interval: 3 6
[3, 6]

The sum of the two intervals is: [5, 14]

The subtraction of the two intervals is: [-4, 5]

The multiplication of the two intervals is: [6, 48]

The reciprocal of the first interval is: [0.5, 0.125]

this is my prog

#include<iostream>
#include<string>

using namespace std;

class interval
{
int lower;
int upper;
void set(int,int)const;
void get (int,int);
void print();
void addtion(int,int,int,int);
void subtract(int,int,int,int);
void mutiplty(int,int,int,int);
void divide(int,int,int,int);
};

int main()
{
interval first,second;
int l,u;

cout<<"Enter the lower and the upper limits of the first interval";
cin>>first.l>>first.u;
cout<<endl;
cout<<"["<<first.l<<","<<first.u<<"]"<<endl;
first.set(l,u);
cout<<"Enter the lower and the upper limits of the second
interval";
cin>>second.l>>second.u;
cout<<endl;
cout<<"["<<second.l<<","<<second.u<<"]"<<endl;

second.set(l,u);
print();

return 0;
}

void interval::set(int,int)const
{
lower=l;
upper=u;
}
void interval::get(int,int)
{
lower=l;
upper=u;
}
void interval::addtion(int,int,int,int)
{ int add1,add2;

add1=first.l+second.l;
add2=first.u+second.u;
}
void interval::subtract(int,int,int,int)
{
int sub1,sub2;
sub1=first.l-second.u;
sub2=first.u-second.l;
}
void interval::mutiplty(int,int,int,int);
{
int ac,ad,bc,bd,mul1,mul2;

ac=first.l*second.l;
ad=first.l*second.u;
bc=first.u*second.l;
bd=first.u*second.u;

mul1=min(ac,ad,bc,bd);
mul2=max(ac,ad,bc,bd);
}
void divide(int,int,int,int);
{
int d1,d2;

if(first.l==0&&first.u==0)
cout<<"error"<<endl;
else
d1=1/first.l;
d2=1/first.u;
}
void print();
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}

Sep 14 '07 #1
10 1712 "CuTe_Engineer" writes:
i have cs assignment i tried to solve it but i still have many
errors , plzz help mee :"< it`s not cheating becuz i`ve tried & wrote
the prog. i just wanna you to show me my mistakes

#these are the operations
[a, b] + [c, d] = [a+c, b+d],
[a, b] - [c, d] = [a-d, b-c],
[a, b] * [c, d] = [min(ac, ad, bc, bd), max(ac, ad, bc, bd)],
1/[a, b] = [1/b, 1/a] only if 0 not in [a,b].

& the Question is to Write the class interval that has lower and upper
as data members, set, get, print, and constructors as member
functions. The class also has the four member functions add, subtract,
multiply, and divide.

Implement the class members, and write a driver that declares two
objects of the class interval and prints the results of the four
operations on the two declared objects.

Sample input / output:

Enter the lower and upper limits of the first interval: 2 8
[2, 8]
Enter the lower and upper limits of the second interval: 3 6
[3, 6]

The sum of the two intervals is: [5, 14]

The subtraction of the two intervals is: [-4, 5]

The multiplication of the two intervals is: [6, 48]

The reciprocal of the first interval is: [0.5, 0.125]

this is my prog

#include<iostream>
#include<string>

using namespace std;

class interval
{
int lower;
int upper;
Private:

That's the first error I note. The lack makes the functions private.
void set(int,int)const;
void get (int,int);
void print();
void addtion(int,int,int,int);
void subtract(int,int,int,int);
void mutiplty(int,int,int,int);
void divide(int,int,int,int);
};

int main()
{
interval first,second;
int l,u;

cout<<"Enter the lower and the upper limits of the first interval";
cin>>first.l>>first.u;
cout<<endl;
cout<<"["<<first.l<<","<<first.u<<"]"<<endl;
first.set(l,u);
cout<<"Enter the lower and the upper limits of the second
interval";
cin>>second.l>>second.u;
cout<<endl;
cout<<"["<<second.l<<","<<second.u<<"]"<<endl;

second.set(l,u);
print();

return 0;
}

void interval::set(int,int)const
{
lower=l;
upper=u;
}
void interval::get(int,int)
{
lower=l;
upper=u;
}
void interval::addtion(int,int,int,int)
{ int add1,add2;

add1=first.l+second.l;
add2=first.u+second.u;
}
void interval::subtract(int,int,int,int)
{
int sub1,sub2;
sub1=first.l-second.u;
sub2=first.u-second.l;
}
void interval::mutiplty(int,int,int,int);
{
int ac,ad,bc,bd,mul1,mul2;

ac=first.l*second.l;
ad=first.l*second.u;
bc=first.u*second.l;
bd=first.u*second.u;

mul1=min(ac,ad,bc,bd);
mul2=max(ac,ad,bc,bd);
}
void divide(int,int,int,int);
{
int d1,d2;

if(first.l==0&&first.u==0)
cout<<"error"<<endl;
else
d1=1/first.l;
d2=1/first.u;
}
void print();
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}

Sep 14 '07 #2
On 2007-09-14 15:06, CuTe_Engineer wrote:
hii,
i have cs assignment i tried to solve it but i still have many
errors , plzz help mee :"< it`s not cheating becuz i`ve tried & wrote
the prog. i just wanna you to show me my mistakes

#these are the operations
[a, b] + [c, d] = [a+c, b+d],
[a, b] - [c, d] = [a-d, b-c],
[a, b] * [c, d] = [min(ac, ad, bc, bd), max(ac, ad, bc, bd)],
1/[a, b] = [1/b, 1/a] only if 0 not in [a,b].

& the Question is to Write the class interval that has lower and upper
as data members, set, get, print, and constructors as member
functions. The class also has the four member functions add, subtract,
multiply, and divide.

Implement the class members, and write a driver that declares two
objects of the class interval and prints the results of the four
operations on the two declared objects.

Sample input / output:

Enter the lower and upper limits of the first interval: 2 8
[2, 8]
Enter the lower and upper limits of the second interval: 3 6
[3, 6]

The sum of the two intervals is: [5, 14]

The subtraction of the two intervals is: [-4, 5]

The multiplication of the two intervals is: [6, 48]

The reciprocal of the first interval is: [0.5, 0.125]

this is my prog

#include<iostream>
#include<string>

using namespace std;

class interval
{
int lower;
int upper;
public:
interval(int l, int u);

Do not forget the constructor.
void set(int,int)const;
Not const, see below for explanation.
void get (int,int);
Will not work, see below.
void print();
void print() const;
void addtion(int,int,int,int);
void add(const interval&) const;
void subtract(int,int,int,int);
void subtract(const interval&) const;
void mutiplty(int,int,int,int);
void multiply(const interval&) const;
void divide(int,int,int,int);
void divide(const interval&) const;
};

int main()
{
interval first,second;
Remove those, you (now) have a constructor, make sure to also use it.
int l,u;

cout<<"Enter the lower and the upper limits of the first interval";
cin >l >u;

interval first(l, u);
cout<<"["<<first.l<<","<<first.u<<"]"<<endl;
first.set(l,u);
No! 1) they are called lower and upper, not l and u, 2) they are
private, you should never change them unless through the usage of member
functions, 3) you should use print() to show their value.
Same goes for this the second one.
print();
print() is a member, you need an object to call it.
>
return 0;
}
interval::interval(int l, int u)
{
// Insert code here, unless you know about initialisation lists in
// which case you should use them.
}
void interval::set(int,int)const
Cannot use const here, it indicates that this function does not make any
changes to the object, which is obviously does.
{
lower=l;
upper=u;
}
void interval::get(int,int)
No, this does not work, you need to pass them as references, or return a
struct containing their values or make get_upper() and get_lower()
functions.

void interval::addtion(int,int,int,int)
You should pass another interval as parameters

void interval::addition(const interval&) const
{ int add1,add2;

add1=first.l+second.l;
add2=first.u+second.u;
}
And you should use the constructor to create the new object.
The same goes for the rest of the functions.
void print();
void interval::print() const
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}
And what are add1 and add2? All those operations should be performed in
main() and it should look something like

cout << "The sum of the two intervals is: ";
(first + second).print();

I hope you have much time left until you need to turn the assignment in
because you have obviously not been paying attention during the lessons
(or your instructor is *really* bad).

--
Erik Wikström
Sep 14 '07 #3

On Sep 14, 8:47 pm, Erik Wikström <Erik-wikst...@telia.comwrote:
On 2007-09-14 15:06, CuTe_Engineer wrote:

hii,
i have cs assignment i tried to solve it but i still have many
errors , plzz help mee :"< it`s not cheating becuz i`ve tried & wrote
the prog. i just wanna you to show me my mistakes
#these are the operations
[a, b] + [c, d] = [a+c, b+d],
[a, b] - [c, d] = [a-d, b-c],
[a, b] * [c, d] = [min(ac, ad, bc, bd), max(ac, ad, bc, bd)],
1/[a, b] = [1/b, 1/a] only if 0 not in [a,b].
& the Question is to Write the class interval that has lower and upper
as data members, set, get, print, and constructors as member
functions. The class also has the four member functions add, subtract,
multiply, and divide.
Implement the class members, and write a driver that declares two
objects of the class interval and prints the results of the four
operations on the two declared objects.
Sample input / output:
Enter the lower and upper limits of the first interval: 2 8
[2, 8]
Enter the lower and upper limits of the second interval: 3 6
[3, 6]
The sum of the two intervals is: [5, 14]
The subtraction of the two intervals is: [-4, 5]
The multiplication of the two intervals is: [6, 48]
The reciprocal of the first interval is: [0.5, 0.125]
this is my prog
#include<iostream>
#include<string>
using namespace std;
class interval
{
int lower;
int upper;

public:
interval(int l, int u);

Do not forget the constructor.
void set(int,int)const;

Not const, see below for explanation.
void get (int,int);

Will not work, see below.
void print();

void print() const;
void addtion(int,int,int,int);

void add(const interval&) const;
void subtract(int,int,int,int);

void subtract(const interval&) const;
void mutiplty(int,int,int,int);

void multiply(const interval&) const;
void divide(int,int,int,int);

void divide(const interval&) const;
};
int main()
{
interval first,second;

Remove those, you (now) have a constructor, make sure to also use it.
int l,u;
cout<<"Enter the lower and the upper limits of the first interval";

cin >l >u;

interval first(l, u);
cout<<"["<<first.l<<","<<first.u<<"]"<<endl;
first.set(l,u);

No! 1) they are called lower and upper, not l and u, 2) they are
private, you should never change them unless through the usage of member
functions, 3) you should use print() to show their value.
Same goes for this the second one.
print();

print() is a member, you need an object to call it.
return 0;
}

interval::interval(int l, int u)
{
// Insert code here, unless you know about initialisation lists in
// which case you should use them.

}
void interval::set(int,int)const

Cannot use const here, it indicates that this function does not make any
changes to the object, which is obviously does.
{
lower=l;
upper=u;
}
void interval::get(int,int)

No, this does not work, you need to pass them as references, or return a
struct containing their values or make get_upper() and get_lower()
functions.
void interval::addtion(int,int,int,int)

You should pass another interval as parameters

void interval::addition(const interval&) const
{ int add1,add2;
add1=first.l+second.l;
add2=first.u+second.u;
}

And you should use the constructor to create the new object.
The same goes for the rest of the functions.
void print();

void interval::print() const
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}

And what are add1 and add2? All those operations should be performed in
main() and it should look something like

cout << "The sum of the two intervals is: ";
(first + second).print();

I hope you have much time left until you need to turn the assignment in
because you have obviously not been paying attention during the lessons
(or your instructor is *really* bad).

--
Erik Wikström- Hide quoted text -

- Show quoted text -
Erik thanks for helping me you are soo kind by the way my teacher is
not *really* bad i like him but maybe there is some thing wrong with
me loOol

ok now i understood your explaination but not everthing :"
can you show me example of how to use the constructor & when should we
use it ? i believe it is used to assgin values to the members in
class ..

void interval::addition(const interval&) const
>
{ int add1,add2;
add1=first.l+second.l;
add2=first.u+second.u;
}

And you should use the constructor to create the new object.
The same goes for the rest of the functions.( can you show me how i don`tknow how to use constuctor*)
void interval::print() const
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}

And what are add1 and add2? All those operations should be performed in
main() and it should look something like

cout << "The sum of the two intervals is: ";
(first + second).print();
if i wrote all these in the main then what should i write in
print!!!
sorry i know that i`m annoying you :""

thhhhhhhhhaaaaaaaaaank you very much for helping me
i should submit my assign on thursday i still have 5 days it`s okey :)

Sep 14 '07 #4
On Sep 15, 4:40 am, CuTe_Engineer <men_3eyoo...@hotmail.comwrote:
On Sep 14, 8:47 pm, Erik Wikström <Erik-wikst...@telia.comwrote:

On 2007-09-14 15:06, CuTe_Engineer wrote:
hii,
i have cs assignment i tried to solve it but i still have many
errors , plzz help mee :"< it`s not cheating becuz i`ve tried & wrote
the prog. i just wanna you to show me my mistakes
#these are the operations
[a, b] + [c, d] = [a+c, b+d],
[a, b] - [c, d] = [a-d, b-c],
[a, b] * [c, d] = [min(ac, ad, bc, bd), max(ac, ad, bc, bd)],
1/[a, b] = [1/b, 1/a] only if 0 not in [a,b].
& the Question is to Write the class interval that has lower and upper
as data members, set, get, print, and constructors as member
functions. The class also has the four member functions add, subtract,
multiply, and divide.
Implement the class members, and write a driver that declares two
objects of the class interval and prints the results of the four
operations on the two declared objects.
Sample input / output:
Enter the lower and upper limits of the first interval: 2 8
[2, 8]
Enter the lower and upper limits of the second interval: 3 6
[3, 6]
The sum of the two intervals is: [5, 14]
The subtraction of the two intervals is: [-4, 5]
The multiplication of the two intervals is: [6, 48]
The reciprocal of the first interval is: [0.5, 0.125]
this is my prog
#include<iostream>
#include<string>
using namespace std;
class interval
{
int lower;
int upper;
public:
interval(int l, int u);
Do not forget the constructor.
void set(int,int)const;
Not const, see below for explanation.
void get (int,int);
Will not work, see below.
void print();
void print() const;
void addtion(int,int,int,int);
void add(const interval&) const;
void subtract(int,int,int,int);
void subtract(const interval&) const;
void mutiplty(int,int,int,int);
void multiply(const interval&) const;
void divide(int,int,int,int);
void divide(const interval&) const;
};
int main()
{
interval first,second;
Remove those, you (now) have a constructor, make sure to also use it.
int l,u;
cout<<"Enter the lower and the upper limits of the first interval";
cin >l >u;
interval first(l, u);
cout<<"["<<first.l<<","<<first.u<<"]"<<endl;
first.set(l,u);
No! 1) they are called lower and upper, not l and u, 2) they are
private, you should never change them unless through the usage of member
functions, 3) you should use print() to show their value.
Same goes for this the second one.
print();
print() is a member, you need an object to call it.
return 0;
}
interval::interval(int l, int u)
{
// Insert code here, unless you know about initialisation lists in
// which case you should use them.
}
void interval::set(int,int)const
Cannot use const here, it indicates that this function does not make any
changes to the object, which is obviously does.
{
lower=l;
upper=u;
}
void interval::get(int,int)
No, this does not work, you need to pass them as references, or return a
struct containing their values or make get_upper() and get_lower()
functions.
void interval::addtion(int,int,int,int)
You should pass another interval as parameters
void interval::addition(const interval&) const
{ int add1,add2;
add1=first.l+second.l;
add2=first.u+second.u;
}
And you should use the constructor to create the new object.
The same goes for the rest of the functions.
void print();
void interval::print() const
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}
And what are add1 and add2? All those operations should be performed in
main() and it should look something like
cout << "The sum of the two intervals is: ";
(first + second).print();
I hope you have much time left until you need to turn the assignment in
because you have obviously not been paying attention during the lessons
(or your instructor is *really* bad).
--
Erik Wikström- Hide quoted text -
- Show quoted text -

Erik thanks for helping me you are soo kind by the way my teacher is
not *really* bad i like him but maybe there is some thing wrong with
me loOol

ok now i understood your explaination but not everthing :"
can you show me example of how to use the constructor & when should we
use it ? i believe it is used to assgin values to the members in
class ..

void interval::addition(const interval&) const

{ int add1,add2;
add1=first.l+second.l;
add2=first.u+second.u;
}
And you should use the constructor to create the new object.
The same goes for the rest of the functions.( can you show me how i don`t know how to use constuctor*)
void interval::print() const
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}
And what are add1 and add2? All those operations should be performed in
main() and it should look something like
cout << "The sum of the two intervals is: ";
(first + second).print();

if i wrote all these in the main then what should i write in
print!!!
sorry i know that i`m annoying you :""

thhhhhhhhhaaaaaaaaaank you very much for helping me
i should submit my assign on thursday i still have 5 days it`s okey :)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -
You should pass another interval as parameters

void interval::addition(const interval&) const <<<<why???

is my operations rigth or wrong???
Sep 14 '07 #5
On 2007-09-14 23:40, CuTe_Engineer wrote:
On Sep 14, 8:47 pm, Erik Wikström <Erik-wikst...@telia.comwrote:
>On 2007-09-14 15:06, CuTe_Engineer wrote:

hii,
i have cs assignment i tried to solve it but i still have many
errors , plzz help mee :"< it`s not cheating becuz i`ve tried & wrote
the prog. i just wanna you to show me my mistakes
#these are the operations
[a, b] + [c, d] = [a+c, b+d],
[a, b] - [c, d] = [a-d, b-c],
[a, b] * [c, d] = [min(ac, ad, bc, bd), max(ac, ad, bc, bd)],
1/[a, b] = [1/b, 1/a] only if 0 not in [a,b].
& the Question is to Write the class interval that has lower and upper
as data members, set, get, print, and constructors as member
functions. The class also has the four member functions add, subtract,
multiply, and divide.
Implement the class members, and write a driver that declares two
objects of the class interval and prints the results of the four
operations on the two declared objects.
Sample input / output:
Enter the lower and upper limits of the first interval: 2 8
[2, 8]
Enter the lower and upper limits of the second interval: 3 6
[3, 6]
The sum of the two intervals is: [5, 14]
The subtraction of the two intervals is: [-4, 5]
The multiplication of the two intervals is: [6, 48]
The reciprocal of the first interval is: [0.5, 0.125]
this is my prog
#include<iostream>
#include<string>
using namespace std;
class interval
{
int lower;
int upper;

public:
interval(int l, int u);

Do not forget the constructor.
void set(int,int)const;

Not const, see below for explanation.
void get (int,int);

Will not work, see below.
void print();

void print() const;
void addtion(int,int,int,int);

void add(const interval&) const;
void subtract(int,int,int,int);

void subtract(const interval&) const;
void mutiplty(int,int,int,int);

void multiply(const interval&) const;
void divide(int,int,int,int);

void divide(const interval&) const;
};
int main()
{
interval first,second;

Remove those, you (now) have a constructor, make sure to also use it.
int l,u;
cout<<"Enter the lower and the upper limits of the first interval";

cin >l >u;

interval first(l, u);
cout<<"["<<first.l<<","<<first.u<<"]"<<endl;
first.set(l,u);

No! 1) they are called lower and upper, not l and u, 2) they are
private, you should never change them unless through the usage of member
functions, 3) you should use print() to show their value.
Same goes for this the second one.
print();

print() is a member, you need an object to call it.
return 0;
}

interval::interval(int l, int u)
{
// Insert code here, unless you know about initialisation lists in
// which case you should use them.

}
void interval::set(int,int)const

Cannot use const here, it indicates that this function does not make any
changes to the object, which is obviously does.
{
lower=l;
upper=u;
}
void interval::get(int,int)

No, this does not work, you need to pass them as references, or return a
struct containing their values or make get_upper() and get_lower()
functions.
void interval::addtion(int,int,int,int)

You should pass another interval as parameters

void interval::addition(const interval&) const
{ int add1,add2;
add1=first.l+second.l;
add2=first.u+second.u;
}

And you should use the constructor to create the new object.
The same goes for the rest of the functions.
void print();

void interval::print() const
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}

And what are add1 and add2? All those operations should be performed in
main() and it should look something like

cout << "The sum of the two intervals is: ";
(first + second).print();

I hope you have much time left until you need to turn the assignment in
because you have obviously not been paying attention during the lessons
(or your instructor is *really* bad).

--
Erik Wikström- Hide quoted text -

- Show quoted text -

Erik thanks for helping me you are soo kind by the way my teacher is
not *really* bad i like him but maybe there is some thing wrong with
me loOol

ok now i understood your explaination but not everthing :"
can you show me example of how to use the constructor & when should we
use it ? i believe it is used to assgin values to the members in
class ..

void interval::addition(const interval&) const
>>
{ int add1,add2;
add1=first.l+second.l;
add2=first.u+second.u;
}

And you should use the constructor to create the new object.
The same goes for the rest of the functions.( can you show me how i don`t know how to use constuctor*)
>void interval::print() const
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}

And what are add1 and add2? All those operations should be performed in
main() and it should look something like

cout << "The sum of the two intervals is: ";
(first + second).print();

if i wrote all these in the main then what should i write in
print!!!
sorry i know that i`m annoying you :""

thhhhhhhhhaaaaaaaaaank you very much for helping me
i should submit my assign on thursday i still have 5 days it`s okey :)
Something that outputs the interval, so if you have code like this

interval ival(1,2);
ival.print();

you should get something like

[1,2]

output on the terminal.

--
Erik Wikström
Sep 14 '07 #6
On 2007-09-14 23:48, CuTe_Engineer wrote:
On Sep 15, 4:40 am, CuTe_Engineer <men_3eyoo...@hotmail.comwrote:
> On Sep 14, 8:47 pm, Erik Wikström <Erik-wikst...@telia.comwrote:

On 2007-09-14 15:06, CuTe_Engineer wrote:
hii,
i have cs assignment i tried to solve it but i still have many
errors , plzz help mee :"< it`s not cheating becuz i`ve tried & wrote
the prog. i just wanna you to show me my mistakes
#these are the operations
[a, b] + [c, d] = [a+c, b+d],
[a, b] - [c, d] = [a-d, b-c],
[a, b] * [c, d] = [min(ac, ad, bc, bd), max(ac, ad, bc, bd)],
1/[a, b] = [1/b, 1/a] only if 0 not in [a,b].
& the Question is to Write the class interval that has lower and upper
as data members, set, get, print, and constructors as member
functions. The class also has the four member functions add, subtract,
multiply, and divide.
Implement the class members, and write a driver that declares two
objects of the class interval and prints the results of the four
operations on the two declared objects.
Sample input / output:
Enter the lower and upper limits of the first interval: 2 8
[2, 8]
Enter the lower and upper limits of the second interval: 3 6
[3, 6]
The sum of the two intervals is: [5, 14]
The subtraction of the two intervals is: [-4, 5]
The multiplication of the two intervals is: [6, 48]
The reciprocal of the first interval is: [0.5, 0.125]
this is my prog
#include<iostream>
#include<string>
using namespace std;
class interval
{
int lower;
int upper;
public:
interval(int l, int u);
Do not forget the constructor.
void set(int,int)const;
Not const, see below for explanation.
void get (int,int);
Will not work, see below.
void print();
void print() const;
void addtion(int,int,int,int);
void add(const interval&) const;
void subtract(int,int,int,int);
void subtract(const interval&) const;
void mutiplty(int,int,int,int);
void multiply(const interval&) const;
void divide(int,int,int,int);
void divide(const interval&) const;
};
int main()
{
interval first,second;
Remove those, you (now) have a constructor, make sure to also use it.
int l,u;
cout<<"Enter the lower and the upper limits of the first interval";
cin >l >u;
interval first(l, u);
cout<<"["<<first.l<<","<<first.u<<"]"<<endl;
first.set(l,u);
No! 1) they are called lower and upper, not l and u, 2) they are
private, you should never change them unless through the usage of member
functions, 3) you should use print() to show their value.
Same goes for this the second one.
print();
print() is a member, you need an object to call it.
return 0;
}
interval::interval(int l, int u)
{
// Insert code here, unless you know about initialisation lists in
// which case you should use them.
}
void interval::set(int,int)const
Cannot use const here, it indicates that this function does not make any
changes to the object, which is obviously does.
{
lower=l;
upper=u;
}
void interval::get(int,int)
No, this does not work, you need to pass them as references, or return a
struct containing their values or make get_upper() and get_lower()
functions.
void interval::addtion(int,int,int,int)
You should pass another interval as parameters
void interval::addition(const interval&) const
{ int add1,add2;
add1=first.l+second.l;
add2=first.u+second.u;
}
And you should use the constructor to create the new object.
The same goes for the rest of the functions.
void print();
void interval::print() const
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}
And what are add1 and add2? All those operations should be performed in
main() and it should look something like
cout << "The sum of the two intervals is: ";
(first + second).print();
I hope you have much time left until you need to turn the assignment in
because you have obviously not been paying attention during the lessons
(or your instructor is *really* bad).
--
Erik Wikström- Hide quoted text -
- Show quoted text -

Erik thanks for helping me you are soo kind by the way my teacher is
not *really* bad i like him but maybe there is some thing wrong with
me loOol

ok now i understood your explaination but not everthing :"
can you show me example of how to use the constructor & when should we
use it ? i believe it is used to assgin values to the members in
class ..

void interval::addition(const interval&) const

{ int add1,add2;
add1=first.l+second.l;
add2=first.u+second.u;
}
And you should use the constructor to create the new object.
The same goes for the rest of the functions.( can you show me how i don`t know how to use constuctor*)
void interval::print() const
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}
And what are add1 and add2? All those operations should be performed in
main() and it should look something like
cout << "The sum of the two intervals is: ";
(first + second).print();

if i wrote all these in the main then what should i write in
print!!!
sorry i know that i`m annoying you :""

thhhhhhhhhaaaaaaaaaank you very much for helping me
i should submit my assign on thursday i still have 5 days it`s okey :)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -
Please don't reply to your own post unless to correct your previous post.
You should pass another interval as parameters

void interval::addition(const interval&) const <<<<why???
To make the operations sane, given that you had made addition a member
function one would assume that the intended use was something like this:

interval a, b;
interval c = a.add(b);

--
Erik Wikström

Sep 14 '07 #7
On 2007-09-14 23:40, CuTe_Engineer wrote:
On Sep 14, 8:47 pm, Erik Wikström <Erik-wikst...@telia.comwrote:
>On 2007-09-14 15:06, CuTe_Engineer wrote:

hii,
i have cs assignment i tried to solve it but i still have many
errors , plzz help mee :"< it`s not cheating becuz i`ve tried & wrote
the prog. i just wanna you to show me my mistakes
#these are the operations
[a, b] + [c, d] = [a+c, b+d],
[a, b] - [c, d] = [a-d, b-c],
[a, b] * [c, d] = [min(ac, ad, bc, bd), max(ac, ad, bc, bd)],
1/[a, b] = [1/b, 1/a] only if 0 not in [a,b].
& the Question is to Write the class interval that has lower and upper
as data members, set, get, print, and constructors as member
functions. The class also has the four member functions add, subtract,
multiply, and divide.
Implement the class members, and write a driver that declares two
objects of the class interval and prints the results of the four
operations on the two declared objects.
Sample input / output:
Enter the lower and upper limits of the first interval: 2 8
[2, 8]
Enter the lower and upper limits of the second interval: 3 6
[3, 6]
The sum of the two intervals is: [5, 14]
The subtraction of the two intervals is: [-4, 5]
The multiplication of the two intervals is: [6, 48]
The reciprocal of the first interval is: [0.5, 0.125]
this is my prog
#include<iostream>
#include<string>
using namespace std;
class interval
{
int lower;
int upper;

public:
interval(int l, int u);

Do not forget the constructor.
void set(int,int)const;

Not const, see below for explanation.
void get (int,int);

Will not work, see below.
void print();

void print() const;
void addtion(int,int,int,int);

void add(const interval&) const;
void subtract(int,int,int,int);

void subtract(const interval&) const;
void mutiplty(int,int,int,int);

void multiply(const interval&) const;
void divide(int,int,int,int);

void divide(const interval&) const;
};
int main()
{
interval first,second;

Remove those, you (now) have a constructor, make sure to also use it.
int l,u;
cout<<"Enter the lower and the upper limits of the first interval";

cin >l >u;

interval first(l, u);
cout<<"["<<first.l<<","<<first.u<<"]"<<endl;
first.set(l,u);

No! 1) they are called lower and upper, not l and u, 2) they are
private, you should never change them unless through the usage of member
functions, 3) you should use print() to show their value.
Same goes for this the second one.
print();

print() is a member, you need an object to call it.
return 0;
}

interval::interval(int l, int u)
{
// Insert code here, unless you know about initialisation lists in
// which case you should use them.

}
void interval::set(int,int)const

Cannot use const here, it indicates that this function does not make any
changes to the object, which is obviously does.
{
lower=l;
upper=u;
}
void interval::get(int,int)

No, this does not work, you need to pass them as references, or return a
struct containing their values or make get_upper() and get_lower()
functions.
void interval::addtion(int,int,int,int)

You should pass another interval as parameters

void interval::addition(const interval&) const
{ int add1,add2;
add1=first.l+second.l;
add2=first.u+second.u;
}

And you should use the constructor to create the new object.
The same goes for the rest of the functions.
void print();

void interval::print() const
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}

And what are add1 and add2? All those operations should be performed in
main() and it should look something like

cout << "The sum of the two intervals is: ";
(first + second).print();

I hope you have much time left until you need to turn the assignment in
because you have obviously not been paying attention during the lessons
(or your instructor is *really* bad).

--
Erik Wikström- Hide quoted text -

- Show quoted text -

Erik thanks for helping me you are soo kind by the way my teacher is
not *really* bad i like him but maybe there is some thing wrong with
me loOol

ok now i understood your explaination but not everthing :"
can you show me example of how to use the constructor & when should we
use it ? i believe it is used to assgin values to the members in
class ..

void interval::addition(const interval&) const
Constructors are used when you create a new object. When an object is
created a constructor will be run, if you have not declared any
constructors the compiler will create a default one for you. Often when
you create an object you know what values you want it to have, so you
can pass those to the constructor. To create an interval between 3 and 7
use the constructor like this:

interval myInterval(3,7);

This is much better (cleaner, more efficient, etc.) than the alternative:

interval myInterval;
myInterval.set(3,7);

--
Erik Wikström
Sep 14 '07 #8
On Sep 15, 5:49 am, Erik Wikström <Erik-wikst...@telia.comwrote:
On 2007-09-14 23:40, CuTe_Engineer wrote:

On Sep 14, 8:47 pm, Erik Wikström <Erik-wikst...@telia.comwrote:
On 2007-09-14 15:06, CuTe_Engineer wrote:
hii,
i have cs assignment i tried to solve it but i still have many
errors , plzz help mee :"< it`s not cheating becuz i`ve tried & wrote
the prog. i just wanna you to show me my mistakes
#these are the operations
[a, b] + [c, d] = [a+c, b+d],
[a, b] - [c, d] = [a-d, b-c],
[a, b] * [c, d] = [min(ac, ad, bc, bd), max(ac, ad, bc, bd)],
1/[a, b] = [1/b, 1/a] only if 0 not in [a,b].
& the Question is to Write the class interval that has lower and upper
as data members, set, get, print, and constructors as member
functions. The class also has the four member functions add, subtract,
multiply, and divide.
Implement the class members, and write a driver that declares two
objects of the class interval and prints the results of the four
operations on the two declared objects.
Sample input / output:
Enter the lower and upper limits of the first interval: 2 8
[2, 8]
Enter the lower and upper limits of the second interval: 3 6
[3, 6]
The sum of the two intervals is: [5, 14]
The subtraction of the two intervals is: [-4, 5]
The multiplication of the two intervals is: [6, 48]
The reciprocal of the first interval is: [0.5, 0.125]
this is my prog
#include<iostream>
#include<string>
using namespace std;
class interval
{
int lower;
int upper;
public:
interval(int l, int u);
Do not forget the constructor.
void set(int,int)const;
Not const, see below for explanation.
void get (int,int);
Will not work, see below.
void print();
void print() const;
void addtion(int,int,int,int);
void add(const interval&) const;
void subtract(int,int,int,int);
void subtract(const interval&) const;
void mutiplty(int,int,int,int);
void multiply(const interval&) const;
void divide(int,int,int,int);
void divide(const interval&) const;
};
int main()
{
interval first,second;
Remove those, you (now) have a constructor, make sure to also use it.
int l,u;
cout<<"Enter the lower and the upper limits of the first interval";
cin >l >u;
interval first(l, u);
cout<<"["<<first.l<<","<<first.u<<"]"<<endl;
first.set(l,u);
No! 1) they are called lower and upper, not l and u, 2) they are
private, you should never change them unless through the usage of member
functions, 3) you should use print() to show their value.
Same goes for this the second one.
print();
print() is a member, you need an object to call it.
return 0;
}
interval::interval(int l, int u)
{
// Insert code here, unless you know about initialisation lists in
// which case you should use them.
}
void interval::set(int,int)const
Cannot use const here, it indicates that this function does not make any
changes to the object, which is obviously does.
{
lower=l;
upper=u;
}
void interval::get(int,int)
No, this does not work, you need to pass them as references, or returna
struct containing their values or make get_upper() and get_lower()
functions.
void interval::addtion(int,int,int,int)
You should pass another interval as parameters
void interval::addition(const interval&) const
{ int add1,add2;
add1=first.l+second.l;
add2=first.u+second.u;
}
And you should use the constructor to create the new object.
The same goes for the rest of the functions.
void print();
void interval::print() const
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}
And what are add1 and add2? All those operations should be performed in
main() and it should look something like
cout << "The sum of the two intervals is: ";
(first + second).print();
I hope you have much time left until you need to turn the assignment in
because you have obviously not been paying attention during the lessons
(or your instructor is *really* bad).
--
Erik Wikström- Hide quoted text -
- Show quoted text -
Erik thanks for helping me you are soo kind by the way my teacher is
not *really* bad i like him but maybe there is some thing wrong with
me loOol
ok now i understood your explaination but not everthing :"
can you show me example of how to use the constructor & when should we
use it ? i believe it is used to assgin values to the members in
class ..
void interval::addition(const interval&) const

Constructors are used when you create a new object. When an object is
created a constructor will be run, if you have not declared any
constructors the compiler will create a default one for you. Often when
you create an object you know what values you want it to have, so you
can pass those to the constructor. To create an interval between 3 and 7
use the constructor like this:

interval myInterval(3,7);

This is much better (cleaner, more efficient, etc.) than the alternative:

interval myInterval;
myInterval.set(3,7);

--
Erik Wikström- Hide quoted text -

- Show quoted text -
this is my new prog. i hope it`s better now
you told that we shouldn`t put all these things in the function print
but i didn`t understand what should we do with it if we didn`t placed
it in print " what`s the benefit of print??"
is my constructor ok now?

sorry i don`t have the software at home to compile it so i can see my
mistakes.

#include<iostream>
#include<string>

using namespace std;

class interval
{
public:
interval();
interval(int ,int );
void set(int,int);
void print() const;
void addittion(const interval&) const;
void subtract(const interval&) const;
void multiply(const interval&) const;
void divide(const interval&) const;
void set(int,int);
void get(int&,int&);
private:
int lower;
int upper;
};

int main()
{
interval interval1,
interval interval2(3,6),

cout<<"Enter the lower and the upper limits of the first interval";
cin>>lower>>upper;
interval1.set(2,8)
cout<<endl;
cout<<"["<<lower<<","<<upper<<"]"<<endl;

cout<<"Enter the lower and the upper limits of the second
interval";
cin>>lower>>upper;
interval2.set(3,6)
cout<<endl;
cout<<"["<<lower<<","<<upper<<"]"<<endl;
interval.print();

return 0;
}
interval::interval()
{
lower=2
upper=8
}
interval::interval(int lower, int upper)
{
lower=3;
upper=6;

void interval::set(int,int)
{
lower=l;
upper=u;
}
void interval::get(int&,int&)const
{
lower=l;
upper=u;
}
void interval::addition(int,int,int,int)
{
int add1,add2;

add1=interval1.lower+interval2.lower;
add2=interval1.upper+interval2.upper;
}
void interval::subtract(int,int,int,int)
{
int sub1,sub2;

sub1=interval1.lower-interval2.upper;
sub2=interval1.upper-interval2.lower;

}
void interval::mutiplty(int,int,int,int);
{
int ac,ad,bc,bd,mul1,mul2;

ac=interval1.lower*interval2.second;
ad=interval1.lower*interval2.upper;
bc=interval1.upper*interval2.lower;
bd=interval1.upper*interval2.upper;

mul1=min(ac,ad,bc,bd);
mul2=max(ac,ad,bc,bd);

}
void divide(int,int,int,int);
{
int d1,d2;

if(interval1.lower==0&&interval1.upper==0)
cout<<"error"<<endl;
else
d1=1/interval1.lower;
d2=1/interval1.upper;

}
void print();
{
cout<<"The sum of the two intervals is:"<<"["<< add1 <<","<< add2
<<"]"<<endl;
cout<<"The subtraction of the two intervals is:"<<"["<< sub1
<<","<< sub2 <<"]";<<endl;
cout<<"The multiplication of the two intervals is:"<<"["<< mul1
<<","<< amul2 <<"]"<<endl;
cout<<"The reciprocal of the first interval is:"<<"["<< d1 <<","<<
d2 <<"]"<<endl;

}
Sep 15 '07 #9
On 2007-09-15 10:05, CuTe_Engineer wrote:
On Sep 15, 5:49 am, Erik Wikström <Erik-wikst...@telia.comwrote:
>On 2007-09-14 23:40, CuTe_Engineer wrote:

On Sep 14, 8:47 pm, Erik Wikström <Erik-wikst...@telia.comwrote:
On 2007-09-14 15:06, CuTe_Engineer wrote:
hii,
i have cs assignment i tried to solve it but i still have many
errors , plzz help mee :"< it`s not cheating becuz i`ve tried & wrote
the prog. i just wanna you to show me my mistakes
#these are the operations
[a, b] + [c, d] = [a+c, b+d],
[a, b] - [c, d] = [a-d, b-c],
[a, b] * [c, d] = [min(ac, ad, bc, bd), max(ac, ad, bc, bd)],
1/[a, b] = [1/b, 1/a] only if 0 not in [a,b].
& the Question is to Write the class interval that has lower and upper
as data members, set, get, print, and constructors as member
functions. The class also has the four member functions add, subtract,
multiply, and divide.
Implement the class members, and write a driver that declares two
objects of the class interval and prints the results of the four
operations on the two declared objects.
Sample input / output:
Enter the lower and upper limits of the first interval: 2 8
[2, 8]
Enter the lower and upper limits of the second interval: 3 6
[3, 6]
The sum of the two intervals is: [5, 14]
The subtraction of the two intervals is: [-4, 5]
The multiplication of the two intervals is: [6, 48]
The reciprocal of the first interval is: [0.5, 0.125]
this is my prog
#include<iostream>
#include<string>
using namespace std;
class interval
{
int lower;
int upper;
>public:
interval(int l, int u);
>Do not forget the constructor.
void set(int,int)const;
>Not const, see below for explanation.
void get (int,int);
>Will not work, see below.
void print();
> void print() const;
void addtion(int,int,int,int);
> void add(const interval&) const;
void subtract(int,int,int,int);
> void subtract(const interval&) const;
void mutiplty(int,int,int,int);
> void multiply(const interval&) const;
void divide(int,int,int,int);
> void divide(const interval&) const;
};
int main()
{
interval first,second;
>Remove those, you (now) have a constructor, make sure to also use it.
int l,u;
cout<<"Enter the lower and the upper limits of the first interval";
> cin >l >u;
> interval first(l, u);
cout<<"["<<first.l<<","<<first.u<<"]"<<endl;
first.set(l,u);
>No! 1) they are called lower and upper, not l and u, 2) they are
private, you should never change them unless through the usage of member
functions, 3) you should use print() to show their value.
Same goes for this the second one.
print();
>print() is a member, you need an object to call it.
return 0;
}
>interval::interval(int l, int u)
{
// Insert code here, unless you know about initialisation lists in
// which case you should use them.
>}
void interval::set(int,int)const
>Cannot use const here, it indicates that this function does not make any
changes to the object, which is obviously does.
{
lower=l;
upper=u;
}
void interval::get(int,int)
>No, this does not work, you need to pass them as references, or return a
struct containing their values or make get_upper() and get_lower()
functions.
void interval::addtion(int,int,int,int)
>You should pass another interval as parameters
>void interval::addition(const interval&) const
{ int add1,add2;
add1=first.l+second.l;
add2=first.u+second.u;
}
>And you should use the constructor to create the new object.
The same goes for the rest of the functions.
void print();
>void interval::print() const
{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}
>And what are add1 and add2? All those operations should be performed in
main() and it should look something like
> cout << "The sum of the two intervals is: ";
(first + second).print();
>I hope you have much time left until you need to turn the assignment in
because you have obviously not been paying attention during the lessons
(or your instructor is *really* bad).
>--
Erik Wikström- Hide quoted text -
>- Show quoted text -
Erik thanks for helping me you are soo kind by the way my teacher is
not *really* bad i like him but maybe there is some thing wrong with
me loOol
ok now i understood your explaination but not everthing :"
can you show me example of how to use the constructor & when should we
use it ? i believe it is used to assgin values to the members in
class ..
void interval::addition(const interval&) const

Constructors are used when you create a new object. When an object is
created a constructor will be run, if you have not declared any
constructors the compiler will create a default one for you. Often when
you create an object you know what values you want it to have, so you
can pass those to the constructor. To create an interval between 3 and 7
use the constructor like this:

interval myInterval(3,7);

This is much better (cleaner, more efficient, etc.) than the alternative:

interval myInterval;
myInterval.set(3,7);

--
Erik Wikström- Hide quoted text -

- Show quoted text -
this is my new prog. i hope it`s better now
you told that we shouldn`t put all these things in the function print
but i didn`t understand what should we do with it if we didn`t placed
it in print " what`s the benefit of print??"
is my constructor ok now?

sorry i don`t have the software at home to compile it so i can see my
mistakes.

#include<iostream>
#include<string>

using namespace std;

class interval
{
public:
interval();
interval(int ,int );
void set(int,int);
void print() const;
void addittion(const interval&) const;
void subtract(const interval&) const;
void multiply(const interval&) const;
void divide(const interval&) const;
void set(int,int);
void get(int&,int&);
private:
int lower;
int upper;
};

int main()
{
interval interval1,
interval interval2(3,6),

cout<<"Enter the lower and the upper limits of the first interval";
cin>>lower>>upper;
interval1.set(2,8)
cout<<endl;
cout<<"["<<lower<<","<<upper<<"]"<<endl;
The idea is that you should read in the values from the user, then pass
those values to the constructor so you do not have to use set(). Then
you will call print() which will do the job of the last line above.

By the way, are you sure you should use predefined values in the
intervals and not those read from the user?
interval::interval()
{
lower=2
upper=8
}
Usually one would use something like 0,0 as default values, or the min
and max values of an integer.
interval::interval(int lower, int upper)
{
lower=3;
upper=6;
}

Perhaps you should make some use of those parameters?
void interval::set(int,int)
{
lower=l;
upper=u;
}
void interval::get(int&,int&)const
{
lower=l;
upper=u;
}
Have you actually tried to compile your code? Where did l and u come from?
void interval::addition(int,int,int,int)
Re-read my previous posts regarding the operations.
void print();
{
cout<<"The sum of the two intervals is:"<<"["<< add1 <<","<< add2
<<"]"<<endl;
cout<<"The subtraction of the two intervals is:"<<"["<< sub1
<<","<< sub2 <<"]";<<endl;
cout<<"The multiplication of the two intervals is:"<<"["<< mul1
<<","<< amul2 <<"]"<<endl;
cout<<"The reciprocal of the first interval is:"<<"["<< d1 <<","<<
d2 <<"]"<<endl;

}
Frankly I do not see much of a difference between this new code and the
original, besides two constructors which you then totally fail to make
use of in any meaningful way.

It is becoming more and more obvious that you have failed to understand
the most basic concepts of procedural programming like scope and
lifetime, until you learn those you can not advance to more advanced
topics like objects (which is a basic concept of object oriented
programming, what you are trying to do). I suggest that you start over
from chapter 1 and read carefully and do the exercises, or talk to your
lecturer about some private tutoring.

--
Erik Wikström
Sep 15 '07 #10

Erik Wikström <Er***********@telia.comwrote in message
news:ok****************@newsb.telia.net...
On 2007-09-15 10:05, CuTe_Engineer wrote:
On Sep 15, 5:49 am, Erik Wikström <Erik-wikst...@telia.comwrote:
On 2007-09-14 23:40, CuTe_Engineer wrote:

On Sep 14, 8:47 pm, Erik Wikström <Erik-wikst...@telia.com>
wrote:
On 2007-09-14 15:06, CuTe_Engineer wrote:

hii,
i have cs assignment i tried to solve it but i still have many
errors , plzz help mee :"< it`s not cheating becuz i`ve tried &
wrote
the prog. i just wanna you to show me my mistakes

#these are the operations
[a, b] + [c, d] = [a+c, b+d],
[a, b] - [c, d] = [a-d, b-c],
[a, b] * [c, d] = [min(ac, ad, bc, bd), max(ac, ad, bc, bd)],
1/[a, b] = [1/b, 1/a] only if 0 not in [a,b].

& the Question is to Write the class interval that has lower and
upper
as data members, set, get, print, and constructors as member
functions. The class also has the four member functions add,
subtract,
multiply, and divide.

Implement the class members, and write a driver that declares two
objects of the class interval and prints the results of the four
operations on the two declared objects.

Sample input / output:

Enter the lower and upper limits of the first interval: 2 8
[2, 8]

Enter the lower and upper limits of the second interval: 3 6
[3, 6]

The sum of the two intervals is: [5, 14]

The subtraction of the two intervals is: [-4, 5]

The multiplication of the two intervals is: [6, 48]

The reciprocal of the first interval is: [0.5, 0.125]

this is my prog

#include<iostream>
#include<string>

using namespace std;

class interval
{
int lower;
int upper;

public:
interval(int l, int u);

Do not forget the constructor.

void set(int,int)const;

Not const, see below for explanation.

void get (int,int);

Will not work, see below.

void print();

void print() const;

void addtion(int,int,int,int);

void add(const interval&) const;

void subtract(int,int,int,int);

void subtract(const interval&) const;

void mutiplty(int,int,int,int);

void multiply(const interval&) const;

void divide(int,int,int,int);

void divide(const interval&) const;

};

int main()
{
interval first,second;

Remove those, you (now) have a constructor, make sure to also use
it.
>
int l,u;

cout<<"Enter the lower and the upper limits of the first
interval";
>
cin >l >u;

interval first(l, u);

cout<<"["<<first.l<<","<<first.u<<"]"<<endl;
first.set(l,u);

No! 1) they are called lower and upper, not l and u, 2) they are
private, you should never change them unless through the usage of
member
functions, 3) you should use print() to show their value.
Same goes for this the second one.

print();

print() is a member, you need an object to call it.

return 0;
}

interval::interval(int l, int u)
{
// Insert code here, unless you know about initialisation lists
in
// which case you should use them.

}
void interval::set(int,int)const

Cannot use const here, it indicates that this function does not make
any
changes to the object, which is obviously does.

{
lower=l;
upper=u;
}
void interval::get(int,int)

No, this does not work, you need to pass them as references, or
return a
struct containing their values or make get_upper() and get_lower()
functions.

void interval::addtion(int,int,int,int)

You should pass another interval as parameters

void interval::addition(const interval&) const

{ int add1,add2;

add1=first.l+second.l;
add2=first.u+second.u;
}

And you should use the constructor to create the new object.
The same goes for the rest of the functions.

void print();

void interval::print() const

{
cout<<"The sum of the two intervals is:
[ "<<add1<<","<<add2<<"]"<<endl;
cout<<"The subtraction of the two intervals is:
[ "<<sub1<<","<<sub2<<"]"<<endl;
cout<<"The mutiplication of the two intervals is:
[ "<<mul1<<","<<mul2<<"]"<<endl;
cout<<"The reciprocal of the first interval is:
[ "<<d1<<","<<d2<<"]"<<endl;
}

And what are add1 and add2? All those operations should be performed
in
main() and it should look something like

cout << "The sum of the two intervals is: ";
(first + second).print();

I hope you have much time left until you need to turn the assignment
in
because you have obviously not been paying attention during the
lessons
(or your instructor is *really* bad).

--
Erik Wikström- Hide quoted text -

- Show quoted text -

Erik thanks for helping me you are soo kind by the way my teacher is
not *really* bad i like him but maybe there is some thing wrong with
me loOol

ok now i understood your explaination but not everthing :"
can you show me example of how to use the constructor & when should
we
use it ? i believe it is used to assgin values to the members in
class ..
void interval::addition(const interval&) const

Constructors are used when you create a new object. When an object is
created a constructor will be run, if you have not declared any
constructors the compiler will create a default one for you. Often when
you create an object you know what values you want it to have, so you
can pass those to the constructor. To create an interval between 3 and
7
use the constructor like this:

interval myInterval(3,7);

This is much better (cleaner, more efficient, etc.) than the
alternative:
>
interval myInterval;
myInterval.set(3,7);

--
Erik Wikström- Hide quoted text -

- Show quoted text -

this is my new prog. i hope it`s better now
you told that we shouldn`t put all these things in the function print
but i didn`t understand what should we do with it if we didn`t placed
it in print " what`s the benefit of print??"
is my constructor ok now?

sorry i don`t have the software at home to compile it so i can see my
mistakes.

#include<iostream>
#include<string>

using namespace std;

class interval
{
public:
interval();
interval(int ,int );
void set(int,int);
void print() const;
void addittion(const interval&) const;
void subtract(const interval&) const;
void multiply(const interval&) const;
void divide(const interval&) const;
void set(int,int);
void get(int&,int&);
private:
int lower;
int upper;
};

int main()
{
interval interval1,
interval interval2(3,6),

cout<<"Enter the lower and the upper limits of the first interval";
cin>>lower>>upper;
interval1.set(2,8)
cout<<endl;
cout<<"["<<lower<<","<<upper<<"]"<<endl;

The idea is that you should read in the values from the user, then pass
those values to the constructor so you do not have to use set(). Then
you will call print() which will do the job of the last line above.

By the way, are you sure you should use predefined values in the
intervals and not those read from the user?
interval::interval()
{
lower=2
upper=8
}

Usually one would use something like 0,0 as default values, or the min
and max values of an integer.
And init lists:

interval::interval() : lower(2), upper(8){}
>
interval::interval(int lower, int upper)
{
lower=3;
upper=6;
}

Perhaps you should make some use of those parameters?
And since....
interval::interval(int lower, int upper){
lower=lower;
upper=upper;
}
..... is confusing(among other things), you should re-name the parms:

interval::interval(int low, int upp){
lower=low;
upper=upp;
}

And use init list:

interval::interval(int low, int upp) : lower(low), upper(upp){}

You (the OP), could combine the two constructors to one:

// interval::interval() : lower(2), upper(8){}
interval::interval( int low = 2, int upp = 8)
: lower(low), upper(upp){}

Since the constructor can now be invoked without parameters, it can be used
as the default constructor ( interval Ival; will work).
>
void interval::set(int,int)
{
lower=l;
upper=u;
}
You (the OP) forgot to name the parms:
void interval::set( int l, int u ){
lower=l;
upper=u;
}
void interval::get(int&,int&)const
{
lower=l;
upper=u;
}

Have you actually tried to compile your code? Where did l and u come from?
void interval::get( int &l, int &u ) const {
// lower=l;
// upper=u;
l = lower;
u = upper;
}

>
void interval::addition(int,int,int,int)

Re-read my previous posts regarding the operations.
void print();
{
cout<<"The sum of the two intervals is:"<<"["<< add1 <<","<< add2
<<"]"<<endl;
cout<<"The subtraction of the two intervals is:"<<"["<< sub1
<<","<< sub2 <<"]";<<endl;
cout<<"The multiplication of the two intervals is:"<<"["<< mul1
<<","<< amul2 <<"]"<<endl;
cout<<"The reciprocal of the first interval is:"<<"["<< d1 <<","<<
d2 <<"]"<<endl;

}

Frankly I do not see much of a difference between this new code and the
original, besides two constructors which you then totally fail to make
use of in any meaningful way.

It is becoming more and more obvious that you have failed to understand
the most basic concepts of procedural programming like scope and
lifetime, until you learn those you can not advance to more advanced
topics like objects (which is a basic concept of object oriented
programming, what you are trying to do). I suggest that you start over
from chapter 1 and read carefully and do the exercises, or talk to your
lecturer about some private tutoring.
- -
Erik Wikström

For OP:

#include<iostream>
#include<string>
// using namespace std;

class interval{ public:
interval( int low = 0, int upp = 0)
: lower( low ), upper( upp ){}
void set(int,int);
void print() const;
void addittion( interval const & ) const;
void subtract( interval const & ) const;
void multiply( interval const & ) const;
void divide( interval const & ) const;
void set( int const low, int const upp){
lower = low;
upper = upp;
return;
}
void get( int &low, int &upp )const{
low = lower;
upp = upper;
return;
}
private:
int lower;
int upper;
};
void print( std::ostream &out, interval const &in1, interval const &in2){
// using namespace std; // if you must
int low1(0), upp1(0), low2(0), upp2(0);
in1.get( low1, upp1 );
in2.get( low2, upp2 );
out<<"interval1 is:["<<low1<<","<<upp1<<"]"<<std::endl;
out<<"interval2 is:["<<low2<<","<<upp2<<"]"<<std::endl;
out<<"The sum of the two intervals is:["
<< low1 + low2 <<","<< upp1 + upp2 <<"]"<<std::endl;
out<<"The subtraction of the two intervals is:["
<< low1 - low2 <<","<< upp1 - upp2 <<"]"<<std::endl;
out<<"The multiplication of the two intervals is:["
<< low1 * low2 <<","<< upp1 * upp2 <<"]"<<std::endl;
// etc....
return;
} // print(ostream&,interval const&,interval const&)
int main(){
// using namespace std; // if you must
interval interval2( 3, 6 );
std::cout<<"Enter the lower and the upper limits of the first
interval";
int low(0), upp(0);
std::cin>>low>>upp;
// if( not low || not upp ){ return EXIT_FAILURE; }
interval interval1( low, upp );
std::cout<<endl;
print( std::cout, interval1, interval2 );
return 0;
} // main()

[ un-tested. may need 'adjusting'. ]

Now, add in your other class member function definitions, *one-by-one*.
Test each time.
Then use those member functions in 'print(....)'. Write a 'print( interval
const &Ival2);' for your class, and test it. (hint: Ival1 will be the class
itself.)
I figure there must be a reason why posts are not being trimmed, so I
didn't.

--
Bob R
POVrookie
Sep 15 '07 #11

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