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Overloading Operators

P: n/a
KL
I am working on a school assignment, so please don't tell me the
solution. I just want some direction.

I am supposed to overload the >, <, ==, !=, >=, and <= operators using
bool. I am having a bit of a problem in seeing what needs to happen
here. I am just not sure how I do this. Will the overloading function
recognize a < and a usual <? Do I do an
IF (a.letters < b.letters){
return true
}
else {
return false
}

?? Is that the logic I need?

KL

Jul 23 '05 #1
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20 Replies


P: n/a
KL wrote:
I am working on a school assignment, so please don't tell me the
solution. I just want some direction.
You are not posting here very often, right? ;-)
I am supposed to overload the >, <, ==, !=, >=, and <= operators using
bool. I am having a bit of a problem in seeing what needs to happen
here.
Well, that depends.
I am just not sure how I do this. Will the overloading function
recognize a < and a usual <?
Not sure what you mean. Your operator is just a function, and you can use
everything in it that you can use in any other function.
Do I do an
IF (a.letters < b.letters){
return true
}
else {
return false
}

?? Is that the logic I need?


Yes, basically. However, you can leave out the if and just write:

return a.letters < b.letters;

since the result of 'a.letters < b.letters' is just a boolean that is true
or false, depending on the values.

Jul 23 '05 #2

P: n/a
KL
Rolf Magnus wrote:
KL wrote:
I am working on a school assignment, so please don't tell me the
solution. I just want some direction.


You are not posting here very often, right? ;-)


That is right. I just post when I get stuck. So basically, not very often.
I am supposed to overload the >, <, ==, !=, >=, and <= operators
using bool. I am having a bit of a problem in seeing what needs to
happen here.


Well, that depends.
I am just not sure how I do this. Will the overloading function
recognize a < and a usual <?


Not sure what you mean. Your operator is just a function, and you can
use everything in it that you can use in any other function.
Do I do an
IF (a.letters < b.letters){
return true
}
else {
return false
}

?? Is that the logic I need?


Yes, basically. However, you can leave out the if and just write:

return a.letters < b.letters;

since the result of 'a.letters < b.letters' is just a boolean that is
true or false, depending on the values.


OK, so when it is called, for example a<b, my parameters (which were already
given) only passes the b. So my question is how to reference the a part.
It could be called anything. How do I reference it in my overloading
function?

Sorry if this seems like a basic question, but just when I think I have that
figured out, it jumps out of my head again.

--

KL

Your argument is sound, nothing but sound. -Benjamin Franklin

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Jul 23 '05 #3

P: n/a
when an statement like (a < b) is invoked, it expands to the following:
a.operator<(b); ie, operator < is invoked for object a passing object b
as the parameter. hence you don't have to worry about object a (
provided a and b are objects of the same type/class/struct/union).that
is also the reason you have only one parameter in the overloaded
function.

Jul 23 '05 #4

P: n/a
"KL" <kl*******@aohell.com> wrote in message
news:42**********@spool9-east.superfeed.net

OK, so when it is called, for example a<b, my parameters (which were
already given) only passes the b. So my question is how to reference
the a part. It could be called anything. How do I reference it in my
overloading function?

Sorry if this seems like a basic question, but just when I think I
have that figured out, it jumps out of my head again.


There are two forms for overloaded operators: member operators and
non-member. Suppose we have an Int class. Using member operators, we might
write it as follows:

class Int
{
int x;
public:
Int(int x_) : x(x_)
{}
bool operator<(const Int& rhs)
{
return x < rhs.x;
}
};

Here we see that the member operator gets the left-hand side of the
inequality expression from its own int member variable. The right-hand side
of the inequality is obtained by accessing the int member from the reference
to Int supplied as an argument. We might use this as follows:

#include <iostream>
int main()
{
Int i1(5);
Int i2(7);
if (i1 < i2)
std::cout << "lhs < rhs is true\n";
else
std::cout << "lhs < rhs is false\n";
}

The disadvantage of member operators is that you can only use them when an
object of the relevant class is on the left-hand side of the operator.
Suppose that you wished to compare Int objects with built-in ints.

int main()
{
Int i1(5);
if (9 < i2)
std::cout << "lhs < rhs is true\n";
else
std::cout << "lhs < rhs is false\n";
}

This won't work because the 9 on the left-hand side is not an Int object and
hence operator< from the Int class cannot be called. You can get around this
by using non-member operators and making them friends. In that case, both
the left-hand side and the right-hand side must be arguments of the
operator.

class Int
{
int x;
public:
Int(int x_) : x(x_)
{}
// Member operator when Int is lhs
// Int-Int comparisons
bool operator<(const Int& rhs)
{
return x < rhs.x;
}
// Int-int comparisons
bool operator<(int rhs)
{
return x < rhs;
}
// Use non-member operator when int is lhs. Make it a friend.
// so it can access private members. Note that it has two parameters.
friend bool operator<(int lhs, const Int& rhs);
};

// define non-member operator
bool operator<(int lhs, const Int &rhs)
{
return lhs < rhs.x;
}

Note that, rather than have a mix of member and non-member operators, many
people prefer to only have non-member operators. In any case, we can now
make all possible comparisons:

int main()
{
Int i1(5);
Int i2(7);
std::cout << "Int-Int comparison\n";
if (i1 < i2)
std::cout << "lhs < rhs is true\n";
else
std::cout << "lhs < rhs is false\n";

std::cout << "int-Int comparison\n";
if (9 < i2)
std::cout << "lhs < rhs is true\n";
else
std::cout << "lhs < rhs is false\n";

std::cout << "Int-int comparison\n";
if (i1 < 9)
std::cout << "lhs < rhs is true\n";
else
std::cout << "lhs < rhs is false\n";
}
--
John Carson

Jul 23 '05 #5

P: n/a
KL
hbsk wrote:
when an statement like (a < b) is invoked, it expands to the
following: a.operator<(b); ie, operator < is invoked for object a
passing object b as the parameter. hence you don't have to worry
about object a ( provided a and b are objects of the same
type/class/struct/union).that is also the reason you have only one
parameter in the overloaded function.


I am still a bit confused. I don't know why I can't grasp this. Here is
what I have so far:
STRING(string s){

int a;

for (a=0; a<100; a++)

letters[a] = ' ';

a = 0;

length = s.length();

while (a < length) {

letters[a] = s.at(a);

a++;}

};

bool operator<(const STRING&) const;

};

bool operator<(const STRING& b) const {

for (int a=0;a<100;a++){

return (s.letters[a]<b.letters[a];

}

}
I get the feeling that this is not completely correct. Any insights?
--

KL

Your argument is sound, nothing but sound. -Benjamin Franklin

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----= East and West-Coast Server Farms - Total Privacy via Encryption =----
Jul 23 '05 #6

P: n/a
KL wrote:
hbsk wrote:
when an statement like (a < b) is invoked, it expands to the
following: a.operator<(b); ie, operator < is invoked for object a
passing object b as the parameter. hence you don't have to worry
about object a ( provided a and b are objects of the same
type/class/struct/union).that is also the reason you have only one
parameter in the overloaded function.

I am still a bit confused. I don't know why I can't grasp this. Here is
what I have so far:


I assume this is part of a definition for a class STRING, but you're
missing the beginning:

class STRING {
STRING(string s){

There's really no reason to pass the ctor parameter s by value as this
will require making a duplicate copy lasting only for the duration of
the constructor execution. Try, STRING (const string& s)
int a;

for (a=0; a<100; a++)

letters[a] = ' ';
Indenting your code will make it easier for others (and you!) to read...

a = 0;

length = s.length();

while (a < length) {

letters[a] = s.at(a);

a++;}
Is letters an array of size 100? If so, this is a risky bit of code if
s is longer than that.
};
If I've counted correctly, the above } closes the ctor definition and
should not be followed by a semicolon.

bool operator<(const STRING&) const;
OK, so you've declared a const member function operator<. Looks good.
};

bool operator<(const STRING& b) const {

for (int a=0;a<100;a++){

return (s.letters[a]<b.letters[a];

}

}
OK, there are a couple problems here. First, your function definition
needs its name fully qualified. This should be:

bool STRING::operator<(const STRING& b) const {

Because the fcn definition is not placed inside the STRING class
definition, the compiler doesn't know that this is a member function of
STRING as opposed to any other class with a < operator.

Fixing that will probably make your code compile (I can't say for sure
since you've left out other parts of the STRING class definition such as
the definition of the array letters) but it probably won't do what you
want it to do. Your for loop will only be executed once because on its
first iteration it encounters a return statement and that's the end of
that. You probably want to check if the characters are equal and if so
move on to the next character before returning a result. Consider aaa <
azz, for example.

Also you need to be careful when comparing characters by value. For
example, it may be that capital Z is less than lowercase a. Is that
what you want? Also I'm not sure about this, but I don't know that
there's any guarantee that a < b < c < ... < z. Perhaps someone who
knows these fine details better than I do can comment on this.


I get the feeling that this is not completely correct. Any insights?


Yeah. See above :)

Hope that helps,
Mark
Jul 23 '05 #7

P: n/a
KL wrote:
hbsk wrote:
when an statement like (a < b) is invoked, it expands to the
following: a.operator<(b); ie, operator < is invoked for object a
passing object b as the parameter. hence you don't have to worry
about object a ( provided a and b are objects of the same
type/class/struct/union).that is also the reason you have only one
parameter in the overloaded function.


I am still a bit confused. I don't know why I can't grasp this. Here is
what I have so far:
STRING(string s){

int a;

for (a=0; a<100; a++)

letters[a] = ' ';

a = 0;

length = s.length();

while (a < length) {

letters[a] = s.at(a);

a++;}

};

bool operator<(const STRING&) const;

};

bool operator<(const STRING& b) const {

for (int a=0;a<100;a++){

return (s.letters[a]<b.letters[a];

}

}
I get the feeling that this is not completely correct. Any insights?


Well, whatever the contents of your STRINGs might be, the for loop is only
run once, because it contains an unconditional return statememt. So what
happens in the loop?

First iteration:
a = 0;
return s.letters[0] < b.letters[0];
Done.

What you probably want for your STRINGs is a lexographical compare. If the
first letter of s is "less" than the first letter of b, you want to return
true. If it's "greater", you want to return false. But if they are equal,
you want to check the second letter and do the same comparison again.
Now try to put the above algorithm in code and try it.
But you're still not done. You have to think about what happens if you reach
the end of one of the strings, since they have probably seldomly both a
lenght of exactly 100 characters. And what happens (what does it mean) if
you reach the end of both of them at the same time?

Jul 23 '05 #8

P: n/a
KL
Rolf Magnus wrote:
KL wrote:
hbsk wrote:
when an statement like (a < b) is invoked, it expands to the
following: a.operator<(b); ie, operator < is invoked for object a
passing object b as the parameter. hence you don't have to worry
about object a ( provided a and b are objects of the same
type/class/struct/union).that is also the reason you have only one
parameter in the overloaded function.
I am still a bit confused. I don't know why I can't grasp this.
Here is what I have so far:
STRING(string s){

int a;

for (a=0; a<100; a++)

letters[a] = ' ';

a = 0;

length = s.length();

while (a < length) {

letters[a] = s.at(a);

a++;}

};

bool operator<(const STRING&) const;

};

bool operator<(const STRING& b) const {

for (int a=0;a<100;a++){

return (s.letters[a]<b.letters[a];

}

}
I get the feeling that this is not completely correct. Any insights?


Well, whatever the contents of your STRINGs might be, the for loop is
only run once, because it contains an unconditional return statememt.
So what happens in the loop?

First iteration:
a = 0;
return s.letters[0] < b.letters[0];
Done.

What you probably want for your STRINGs is a lexographical compare.
If the first letter of s is "less" than the first letter of b, you
want to return true. If it's "greater", you want to return false. But
if they are equal, you want to check the second letter and do the
same comparison again.


OK will think on this once I fully wake up ;-)
Now try to put the above algorithm in code and try it.
But you're still not done. You have to think about what happens if
you reach the end of one of the strings, since they have probably
seldomly both a lenght of exactly 100 characters. And what happens
(what does it mean) if you reach the end of both of them at the same
time?


I was thinking it wouldn't matter for this excercise, because STRING is
loaded with spaces before it gets data. So.....I can just run it for 100
characters, because once the data is done, the rest will be spaces. Am I ok
in this line of thinking?

--

KL

Your argument is sound, nothing but sound. -Benjamin Franklin

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----= East and West-Coast Server Farms - Total Privacy via Encryption =----
Jul 23 '05 #9

P: n/a
KL
KL wrote:
Rolf Magnus wrote:
KL wrote:
hbsk wrote:
when an statement like (a < b) is invoked, it expands to the
following: a.operator<(b); ie, operator < is invoked for object a
passing object b as the parameter. hence you don't have to worry
about object a ( provided a and b are objects of the same
type/class/struct/union).that is also the reason you have only one
parameter in the overloaded function.

I am still a bit confused. I don't know why I can't grasp this.
Here is what I have so far:
STRING(string s){

int a;

for (a=0; a<100; a++)

letters[a] = ' ';

a = 0;

length = s.length();

while (a < length) {

letters[a] = s.at(a);

a++;}

};

bool operator<(const STRING&) const;

};

bool operator<(const STRING& b) const {

for (int a=0;a<100;a++){

return (s.letters[a]<b.letters[a];

}

}
I get the feeling that this is not completely correct. Any
insights?


Well, whatever the contents of your STRINGs might be, the for loop is
only run once, because it contains an unconditional return statememt.
So what happens in the loop?

First iteration:
a = 0;
return s.letters[0] < b.letters[0];
Done.

What you probably want for your STRINGs is a lexographical compare.
If the first letter of s is "less" than the first letter of b, you
want to return true. If it's "greater", you want to return false. But
if they are equal, you want to check the second letter and do the
same comparison again.


OK will think on this once I fully wake up ;-)
Now try to put the above algorithm in code and try it.
But you're still not done. You have to think about what happens if
you reach the end of one of the strings, since they have probably
seldomly both a lenght of exactly 100 characters. And what happens
(what does it mean) if you reach the end of both of them at the same
time?


I was thinking it wouldn't matter for this excercise, because STRING
is loaded with spaces before it gets data. So.....I can just run it
for 100 characters, because once the data is done, the rest will be
spaces. Am I ok in this line of thinking?


Ixnay that last bit! I think I got it figured out....mostly. Problem is I
get a true value returned no matter what. Could you look at this and help
me see where my reasoning ran out the door?

bool operator<(const STRING& b) const{

for (int a=0;a<100;a++){

if( letters[a] != b.letters[a]){

return (letters[a]<b.letters[a]);

}

}

};

--

KL

Your argument is sound, nothing but sound. -Benjamin Franklin

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----
Jul 23 '05 #10

P: n/a
KL wrote:
KL wrote:
Rolf Magnus wrote:
KL wrote:
hbsk wrote:

>when an statement like (a < b) is invoked, it expands to the
>following: a.operator<(b); ie, operator < is invoked for object a
>passing object b as the parameter. hence you don't have to worry
>about object a ( provided a and b are objects of the same
>type/class/struct/union).that is also the reason you have only one
>parameter in the overloaded function.

I am still a bit confused. I don't know why I can't grasp this.
Here is what I have so far:
STRING(string s){

int a;

for (a=0; a<100; a++)

letters[a] = ' ';

a = 0;

length = s.length();

while (a < length) {

letters[a] = s.at(a);

a++;}

};

bool operator<(const STRING&) const;

};

bool operator<(const STRING& b) const {

for (int a=0;a<100;a++){

return (s.letters[a]<b.letters[a];

}

}
I get the feeling that this is not completely correct. Any
insights?

Well, whatever the contents of your STRINGs might be, the for loop is
only run once, because it contains an unconditional return statememt.
So what happens in the loop?

First iteration:
a = 0;
return s.letters[0] < b.letters[0];
Done.

What you probably want for your STRINGs is a lexographical compare.
If the first letter of s is "less" than the first letter of b, you
want to return true. If it's "greater", you want to return false. But
if they are equal, you want to check the second letter and do the
same comparison again.


OK will think on this once I fully wake up ;-)

Now try to put the above algorithm in code and try it.
But you're still not done. You have to think about what happens if
you reach the end of one of the strings, since they have probably
seldomly both a lenght of exactly 100 characters. And what happens
(what does it mean) if you reach the end of both of them at the same
time?


I was thinking it wouldn't matter for this excercise, because STRING
is loaded with spaces before it gets data. So.....I can just run it
for 100 characters, because once the data is done, the rest will be
spaces. Am I ok in this line of thinking?

Ixnay that last bit! I think I got it figured out....mostly. Problem is I
get a true value returned no matter what. Could you look at this and help
me see where my reasoning ran out the door?

bool operator<(const STRING& b) const{

for (int a=0;a<100;a++){

if( letters[a] != b.letters[a]){

return (letters[a]<b.letters[a]);

}

}

};

I don't see why this whould always return true, but it still has a
problem. What happens if the two strings are equal? What (if anything,
*hint*) gets returned?
Jul 23 '05 #11

P: n/a
KL
Mark P wrote:
KL wrote:
KL wrote:
Rolf Magnus wrote:

KL wrote:
> hbsk wrote:
>
>> when an statement like (a < b) is invoked, it expands to the
>> following: a.operator<(b); ie, operator < is invoked for object a
>> passing object b as the parameter. hence you don't have to worry
>> about object a ( provided a and b are objects of the same
>> type/class/struct/union).that is also the reason you have only
>> one parameter in the overloaded function.
>
> I am still a bit confused. I don't know why I can't grasp this.
> Here is what I have so far:
> STRING(string s){
>
> int a;
>
> for (a=0; a<100; a++)
>
> letters[a] = ' ';
>
> a = 0;
>
> length = s.length();
>
> while (a < length) {
>
> letters[a] = s.at(a);
>
> a++;}
>
> };
>
> bool operator<(const STRING&) const;
>
> };
>
> bool operator<(const STRING& b) const {
>
> for (int a=0;a<100;a++){
>
> return (s.letters[a]<b.letters[a];
>
> }
>
> }
>
>
> I get the feeling that this is not completely correct. Any
> insights?

Well, whatever the contents of your STRINGs might be, the for loop
is only run once, because it contains an unconditional return
statememt. So what happens in the loop?

First iteration:
a = 0;
return s.letters[0] < b.letters[0];
Done.

What you probably want for your STRINGs is a lexographical compare.
If the first letter of s is "less" than the first letter of b, you
want to return true. If it's "greater", you want to return false.
But if they are equal, you want to check the second letter and do
the same comparison again.

OK will think on this once I fully wake up ;-)
Now try to put the above algorithm in code and try it.
But you're still not done. You have to think about what happens if
you reach the end of one of the strings, since they have probably
seldomly both a lenght of exactly 100 characters. And what happens
(what does it mean) if you reach the end of both of them at the
same time?

I was thinking it wouldn't matter for this excercise, because STRING
is loaded with spaces before it gets data. So.....I can just run it
for 100 characters, because once the data is done, the rest will be
spaces. Am I ok in this line of thinking?

Ixnay that last bit! I think I got it figured out....mostly. Problem is
I get a true value returned no matter what. Could you
look at this and help me see where my reasoning ran out the door?

bool operator<(const STRING& b) const{

for (int a=0;a<100;a++){

if( letters[a] != b.letters[a]){

return (letters[a]<b.letters[a]);

}

}

};

I don't see why this whould always return true, but it still has a
problem. What happens if the two strings are equal? What (if
anything, *hint*) gets returned?


DOH! It is always the little things.....and I don't get why I always get a
true value, but I do...gonna ponder on that some more and work on a
different return for ....... uh oh....it is a bool, it can only return true
or false. So I can't test for strings that are equal....or can I?

--

KL

Your argument is sound, nothing but sound. -Benjamin Franklin

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----
Jul 23 '05 #12

P: n/a
KL wrote:
I don't see why this whould always return true, but it still has a
problem. What happens if the two strings are equal? What (if
anything, *hint*) gets returned?


DOH! It is always the little things.....and I don't get why I always get
a true value, but I do...gonna ponder on that some more and work on a
different return for ....... uh oh....it is a bool, it can only return
true or false. So I can't test for strings that are equal....or can I?


The question was what your operator returns in the case of both being equal.
How often does your loop iterate? What happens as last thing? What is
returned? What is it supposed to return?

And btw: You cannot test the strings for equality with one call to your
operator<, but with two, you can.
Jul 23 '05 #13

P: n/a
KL
Rolf Magnus wrote:
KL wrote:
I don't see why this whould always return true, but it still has a
problem. What happens if the two strings are equal? What (if
anything, *hint*) gets returned?
DOH! It is always the little things.....and I don't get why I
always get a true value, but I do...gonna ponder on that some more
and work on a different return for ....... uh oh....it is a bool, it
can only return true or false. So I can't test for strings that are
equal....or can I?


The question was what your operator returns in the case of both being
equal. How often does your loop iterate? What happens as last thing?
What is returned? What is it supposed to return?


If the letters are equal then it increments by one and tests for equality of
the next two letters, once there are letters that aren't equal then it tests
for inequality. This loops thru until an inequality is found, or 100
characters are read. My problem seems to be if they strings are identical,
and the fact that I always get a true statement.
And btw: You cannot test the strings for equality with one call to
your operator<, but with two, you can.


I don't understand why I can't test for equality the way I have it?

--

KL

Your argument is sound, nothing but sound. -Benjamin Franklin

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Jul 23 '05 #14

P: n/a
KL wrote:
Mark P wrote:

[snip]

I don't see why this whould always return true, but it still has a
problem. What happens if the two strings are equal? What (if
anything, *hint*) gets returned?

DOH! It is always the little things.....and I don't get why I always get a
true value, but I do...gonna ponder on that some more and work on a
different return for ....... uh oh....it is a bool, it can only return true
or false. So I can't test for strings that are equal....or can I?


As to why you always get "true", it would be helpful if you'd post a
complete, compiling section of code that exhibits the problem.

As to other issue I mentioned, it's not that there's no proper return
value. If two strings are equal, it's usually sensible for < to return
false since the one is *not* less than the other. The problem in your
case is that nothing gets returned because the return statement is never
executed.
Jul 23 '05 #15

P: n/a
KL
Mark P wrote:
KL wrote:
Mark P wrote:

[snip]

I don't see why this whould always return true, but it still has a
problem. What happens if the two strings are equal? What (if
anything, *hint*) gets returned?

DOH! It is always the little things.....and I don't get why I
always get a true value, but I do...gonna ponder on that some more
and work on a different return for ....... uh oh....it is a bool, it
can only return true or false. So I can't test for strings that are
equal....or can I?


As to why you always get "true", it would be helpful if you'd post a
complete, compiling section of code that exhibits the problem.

As to other issue I mentioned, it's not that there's no proper return
value. If two strings are equal, it's usually sensible for < to
return false since the one is *not* less than the other. The problem
in your case is that nothing gets returned because the return
statement is never executed.

class STRING {

char letters[100];

int length;

public:

STRING( ){

length = 0;

for (int a=0; a<100; a++)

letters[a] = ' ';

}

STRING(int num);

STRING(string s){

int a;

for (a=0; a<100; a++)

letters[a] = ' ';

a = 0;

length = s.length();

while (a < length) {

letters[a] = s.at(a);

a++;

}

};

bool operator<(const STRING& b) const{

for (int a=0;a<100;a++){

if( letters[a] != b.letters[a]){

return (letters[a]<b.letters[a]);

}

}

};

friend STRING operator+(const STRING&, const STRING&);

friend int substring(const STRING&, const STRING&);

friend STRING insert(const STRING&, const STRING&, int n);

friend STRING delet(const STRING&, int n, int x);

friend ostream& operator<<(ostream&, const STRING&);

friend istream& operator>>(istream&, STRING&);

};

ostream& operator<< (ostream& os, const STRING& str) {

int a = 0;

while (a < str.length)

os << str.letters[a++];

return os;

}

istream& operator>> (istream& is, STRING& str) {

string s;

getline(is, s);

str.length = s.length();

if (str.length > 100) str.length = 100;

for (int a=0; a<str.length; a++)

str.letters[a] = s.at(a);

return is;

}

STRING operator+(const STRING& a, const STRING& b) {

STRING c,temp_a;

int index=a.length;

int result_length;

int temp_b;

temp_a=a;

result_length=a.length + b.length;

if (result_length > 100){

result_length = 100;

}

temp_b=0;

while (index < result_length){

temp_a.letters[index] = b.letters[temp_b];

temp_b+=1;

index+=1;

}

temp_a.length=result_length;

c=temp_a;

return c;

}

int substring (const STRING& source, const STRING& search){

int num = search.length, pt=0, h=0, g=0,index=0,x;

STRING temp;

while (h <= (source.length - num )) {

index=0;

while (index < search.length){

temp.letters[index] = source.letters[h+index];

index++;

}

h++;

x=0;

g=0;

while (g < num){

if (temp.letters[g] == search.letters[g]){

x++;

g++;

}

else

g=num;

}

if (x==num)

{

pt=h-1;

return pt;

}

}

return -1;

}

STRING insert( const STRING& h, const STRING& i, int n){

STRING temps;

int stringsizes=h.length + i.length, k = 0, x=0 ,l = 0, z =0, j=0,

il=i.length;

if (n>h.length)

stringsizes=stringsizes+n-h.length;

if (stringsizes>100){

stringsizes=100;

}

temps.length=stringsizes;

while (j< n){

temps.letters[j]=h.letters[j];

j++;

}

while (x<il){

temps.letters[j]=i.letters[x];

j++;

x++;

}

while (j<stringsizes){

temps.letters[j] = h.letters[n];

n++;

j++;

}

return temps;

}

STRING delet(const STRING& a, int n, int x){

int count=0, d=0;

STRING f;

f.length=a.length;

while (count<n){

f.letters[count]=a.letters[count];

count++;

}

d=(n+x)+1;

while (d<100){

f.letters[count]=a.letters[d];

count++;

d++;

}

return f;

}

int main( ) {

STRING a, b;

bool n;

string c;

cout << "Enter a string: ";

cin >> a;

cout << "Enter a second string: ";

cin >> b;

n = a<b;

if (n=true){

c="The first string is less than the second string";

}

else {

c="The second string is less that the first string";

}

cout << endl<<n << endl;

cout << c << endl;
return 0;

}
OK this is what I got so far. It is still giving me true every time. I
thought I had that logic ok. I open the for loop, then the if != and return
true if a<b, otherwise I close the if loop and continue the for loop. Am I
totally lost? Do I possibly need a return after the end of the for loop?
Shouldn't it return false if a>b?
--

KL

Your argument is sound, nothing but sound. -Benjamin Franklin

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Jul 23 '05 #16

P: n/a
KL wrote:
int main( ) {

STRING a, b;

bool n;

string c;

cout << "Enter a string: ";

cin >> a;

cout << "Enter a second string: ";

cin >> b;

n = a<b;

if (n=true){
Look at that condition again.
c="The first string is less than the second string";

}

else {

c="The second string is less that the first string";

}

cout << endl<<n << endl;

cout << c << endl;
return 0;

}
OK this is what I got so far. It is still giving me true every time. I
thought I had that logic ok. I open the for loop, then the if != and
return true if a<b, otherwise I close the if loop and continue the for
loop. Am I totally lost?
No, the logic is ok, except for the case where both strings are fully equal.
The reason you seem to get true every time is not a fault in your
operator<, but a fault in your main() function. See above.
Do I possibly need a return after the end of the for loop?
Yes.
Shouldn't it return false if a>b?


It probably does.

Jul 23 '05 #17

P: n/a
KL
Rolf Magnus wrote:
KL wrote:
int main( ) {

STRING a, b;

bool n;

string c;

cout << "Enter a string: ";

cin >> a;

cout << "Enter a second string: ";

cin >> b;

n = a<b;

if (n=true){
Look at that condition again.


Ohhhhh right it returns a 0 or a 1 as a boolean, right? God, booleans will
be the death of me...among other C++ errors I make repeatedly!
c="The first string is less than the second string";

}

else {

c="The second string is less that the first string";

}

cout << endl<<n << endl;

cout << c << endl;
return 0;

}
OK this is what I got so far. It is still giving me true every
time. I thought I had that logic ok. I open the for loop, then the
if != and return true if a<b, otherwise I close the if loop and
continue the for loop. Am I totally lost?
No, the logic is ok, except for the case where both strings are fully
equal. The reason you seem to get true every time is not a fault in
your operator<, but a fault in your main() function. See above.
Do I possibly need a return after the end of the for loop?


Yes.

Now I must figure out what it can return besides 1 or 0. Going to think on
that.....
Shouldn't it return false if a>b?


It probably does.


--

KL

Your argument is sound, nothing but sound. -Benjamin Franklin

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Jul 23 '05 #18

P: n/a
KL
Rolf Magnus wrote:
KL wrote:
int main( ) {

STRING a, b;

bool n;

string c;

cout << "Enter a string: ";

cin >> a;

cout << "Enter a second string: ";

cin >> b;

n = a<b;

if (n=true){


Look at that condition again.
c="The first string is less than the second string";

}

else {

c="The second string is less that the first string";

}

cout << endl<<n << endl;

cout << c << endl;
return 0;

}
OK this is what I got so far. It is still giving me true every
time. I thought I had that logic ok. I open the for loop, then the
if != and return true if a<b, otherwise I close the if loop and
continue the for loop. Am I totally lost?


No, the logic is ok, except for the case where both strings are fully
equal. The reason you seem to get true every time is not a fault in
your operator<, but a fault in your main() function. See above.
Do I possibly need a return after the end of the for loop?


Yes.
Shouldn't it return false if a>b?


It probably does.


OMG! The == !!! I told ya I keep making the same mistakes! That is just
one of them.

OK so that works ok as long as the STRINGs aren't equal. Since we were told
to write this as a bool, I am going to forget about if it is equal. Will
check with the teacher on that though.

Thanks for your help! I am sorry I was being so dense. It's been a VERY
long week here.

--

KL

Your argument is sound, nothing but sound. -Benjamin Franklin

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Jul 23 '05 #19

P: n/a
KL wrote:
OK this is what I got so far. It is still giving me true every
time. I thought I had that logic ok. I open the for loop, then the
if != and return true if a<b, otherwise I close the if loop and
continue the for loop. Am I totally lost?
No, the logic is ok, except for the case where both strings are fully
equal. The reason you seem to get true every time is not a fault in
your operator<, but a fault in your main() function. See above.
Do I possibly need a return after the end of the for loop?


Yes.
Shouldn't it return false if a>b?


It probably does.


OMG! The == !!! I told ya I keep making the same mistakes! That is just
one of them.


I was hoping that you find that one yourself. It makes the learning
experience stronger. ;-)
OK so that works ok as long as the STRINGs aren't equal. Since we were
told to write this as a bool, I am going to forget about if it is equal.
No, it's not something like a 3rd state you need to return. Of course, you
stick with bool, which only has two possible values, true and false.
But if your strings are equal, the first is not less than the second one. So
your operator should definitely return false. What does it do instead?
Will check with the teacher on that though.

Thanks for your help! I am sorry I was being so dense. It's been a VERY
long week here.


Actually, it was fun giving you hints, but still letting you work it out
yourself. :-)

Jul 23 '05 #20

P: n/a
KL
Rolf Magnus wrote:
KL wrote:
OK this is what I got so far. It is still giving me true every
time. I thought I had that logic ok. I open the for loop, then
the if != and return true if a<b, otherwise I close the if loop and
continue the for loop. Am I totally lost?

No, the logic is ok, except for the case where both strings are
fully equal. The reason you seem to get true every time is not a
fault in your operator<, but a fault in your main() function. See
above.

Do I possibly need a return after the end of the for loop?

Yes.

Shouldn't it return false if a>b?

It probably does.


OMG! The == !!! I told ya I keep making the same mistakes! That
is just one of them.


I was hoping that you find that one yourself. It makes the learning
experience stronger. ;-)


Yes it does! And I constantly do that. I think taking Calculus while
learning something like C++ with == in it is just wrong! LOL
OK so that works ok as long as the STRINGs aren't equal. Since we
were told to write this as a bool, I am going to forget about if it
is equal.


No, it's not something like a 3rd state you need to return. Of
course, you stick with bool, which only has two possible values, true
and false.
But if your strings are equal, the first is not less than the second
one. So your operator should definitely return false. What does it do
instead?


Yeah, I figured this out. Was lost, but now I am found!
Will check with the teacher on that though.

Thanks for your help! I am sorry I was being so dense. It's been a
VERY long week here.


Actually, it was fun giving you hints, but still letting you work it
out yourself. :-)


Well, I am sure I will be back when we get our next project. My mind is
numb at this point!

--

KL

Your argument is sound, nothing but sound. -Benjamin Franklin

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Jul 23 '05 #21

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