Right triangles...

[deanfamily11]

Now, I've made a program that when the user enters 3 numbers, it is supposed
to tell the user if it is a right triangle by using the quadratic equation
(a^2 + b^2 = c^2). Now, granted it can be difficult to come up with 3
numbers on the fly that when run through that equation come out true, but
even with using 0's, it should be correct. If there is something wrong with
the following code, I'd appreciate the help, otherwise, tell me some numbers
that will work.
#include <iomanip>
#include <iostream>

using namespace std;

int main()
{

//declare variables
int side1, side2, side3;
double sum;

//get sides from the user
cout << "Enter 3 integers representing sides of a triangle: ";
cin >> side1 >> side2 >> side3;

//determine if it is a right triangle
sum = (side1 ^ 2) + (side2 ^ 2);
side3 = (side3 ^ 2);

if (sum == side3)
cout << "This is a right triangle.";
else
cout << "This is not a right triangle.";

return 0;

}

[GB]

deanfamily11 wrote:

the following code, I'd appreciate the help, otherwise, tell me some numbers
that will work.
side3 = (side3 ^ 2);

The "^" operator does not perform exponentiation. It performs bitwise
exclusive or.

Gregg

[Dave Townsend]

the numbers 3 4 5 and multiple of them will work.
You also need to test all the combinations of the three numbners,
that is, if I enter 5 3 4, you program won't detect a right angle triangle.
"deanfamily 11" <de**********@v erizon.net> wrote in message
news:QKrUe.165$ R9.0@trnddc02.. .

Now, I've made a program that when the user enters 3 numbers, it is supposed to tell the user if it is a right triangle by using the quadratic equation
(a^2 + b^2 = c^2). Now, granted it can be difficult to come up with 3
numbers on the fly that when run through that equation come out true, but
even with using 0's, it should be correct. If there is something wrong with the following code, I'd appreciate the help, otherwise, tell me some numbers that will work.
#include <iomanip>
#include <iostream>

using namespace std;

int main()
{

//declare variables
int side1, side2, side3;
double sum;

//get sides from the user
cout << "Enter 3 integers representing sides of a triangle: ";
cin >> side1 >> side2 >> side3;

//determine if it is a right triangle
sum = (side1 ^ 2) + (side2 ^ 2);
side3 = (side3 ^ 2);

if (sum == side3)
cout << "This is a right triangle.";
else
cout << "This is not a right triangle.";

return 0;

}

[M]

Yes, the code assumes that the third number entered is the hypotenuse
of the triangle. (If the sides need to be entered in any order, the
code could just use the largest of the three values as the
hypotenuse.)

But the bigger problem is that in C, the '^' character is not used for
exponentiation; it's the bitwise exclusive or operator.
On Fri, 9 Sep 2005 21:39:01 -0700, "Dave Townsend"
<da********@com cast.net> wrote:

the numbers 3 4 5 and multiple of them will work.
You also need to test all the combinations of the three numbners,
that is, if I enter 5 3 4, you program won't detect a right angle triangle.
"deanfamily1 1" <de**********@v erizon.net> wrote in message
news:QKrUe.165 $R9.0@trnddc02. ..

Now, I've made a program that when the user enters 3 numbers, it is

supposed

to tell the user if it is a right triangle by using the quadratic equation
(a^2 + b^2 = c^2). Now, granted it can be difficult to come up with 3
numbers on the fly that when run through that equation come out true, but
even with using 0's, it should be correct. If there is something wrong

with

the following code, I'd appreciate the help, otherwise, tell me some

numbers

that will work.
#include <iomanip>
#include <iostream>

using namespace std;

int main()
{

//declare variables
int side1, side2, side3;
double sum;

//get sides from the user
cout << "Enter 3 integers representing sides of a triangle: ";
cin >> side1 >> side2 >> side3;

//determine if it is a right triangle
sum = (side1 ^ 2) + (side2 ^ 2);
side3 = (side3 ^ 2);

if (sum == side3)
cout << "This is a right triangle.";
else
cout << "This is not a right triangle.";

return 0;

}

[Oliver \(Nospam\)]

I inserted corrections below:

"deanfamily 11" <de**********@v erizon.net> schrieb im Newsbeitrag
news:QKrUe.165$ R9.0@trnddc02.. .

#include <iomanip>
#include <iostream>

using namespace std;

int main()
{

//declare variables
int side1, side2, side3;
double sum;

//get sides from the user
cout << "Enter 3 integers representing sides of a triangle: ";
cin >> side1 >> side2 >> side3;

//determine if it is a right triangle
sum = (side1 ^ 2) + (side2 ^ 2);
"^" is taken from basic. In C you call the function "pow":
sum = pow(side1, 2) + pow(side2, 2);
side3 = (side3 ^ 2);
side3 = pow(side3, 2);

if (sum == side3)
Because floating numbers are never as precise, that equality can always be
guaranteed, even in case were equality is given, you should write
if ( fabs(sum-side3) < 0.01 )
cout << "This is a right triangle.";
else
cout << "This is not a right triangle.";

return 0;

}

[John Carson]

"Oliver (Nospam)" <ok*****@arcor. de> wrote in message
news:43******** **************@ newsread2.arcor-online.net

I inserted corrections below:

"deanfamily 11" <de**********@v erizon.net> schrieb im Newsbeitrag
news:QKrUe.165$ R9.0@trnddc02.. .

#include <iomanip>
#include <iostream>

using namespace std;

int main()
{

//declare variables
int side1, side2, side3;
double sum;

//get sides from the user
cout << "Enter 3 integers representing sides of a triangle: ";
cin >> side1 >> side2 >> side3;

//determine if it is a right triangle
sum = (side1 ^ 2) + (side2 ^ 2);

"^" is taken from basic. In C you call the function "pow":
sum = pow(side1, 2) + pow(side2, 2);

side3 = (side3 ^ 2);

side3 = pow(side3, 2);

if (sum == side3)

Because floating numbers are never as precise, that equality can
always be guaranteed, even in case were equality is given, you should
write
if ( fabs(sum-side3) < 0.01 )

cout << "This is a right triangle.";
else
cout << "This is not a right triangle.";

return 0;

}

Given that the user is entering integers, you can test exactly (provided the
numbers are not too big) with:

sum = side1*side1 + side2*side2;
side3 = side3*side3;

if(sum == side3)
//

--
John Carson

[Oliver \(Nospam\)]

Yeap, you are wright, but look at the original source code: "double sum;"

// oliver

[Kai-Uwe Bux]

deanfamily11 wrote:

Now, I've made a program that when the user enters 3 numbers, it is
supposed to tell the user if it is a right triangle by using the quadratic
equation
(a^2 + b^2 = c^2). Now, granted it can be difficult to come up with 3
numbers on the fly that when run through that equation come out true, but
even with using 0's, it should be correct. If there is something wrong
with the following code, I'd appreciate the help, otherwise, tell me some
numbers that will work.
#include <iomanip>
#include <iostream>

using namespace std;

int main()
{

//declare variables
int side1, side2, side3;
double sum;

//get sides from the user
cout << "Enter 3 integers representing sides of a triangle: ";
cin >> side1 >> side2 >> side3;

//determine if it is a right triangle
sum = (side1 ^ 2) + (side2 ^ 2);
side3 = (side3 ^ 2);

if (sum == side3)
cout << "This is a right triangle.";
else
cout << "This is not a right triangle.";

return 0;

}

Since others have already pointed out the bit about ^, I will just remark
that fixing that won't do. Consider

#include <iomanip>
#include <iostream>

using namespace std;

int main()
{

//declare variables
int side1, side2, side3;
double sum;

//get sides from the user
cout << "Enter 3 integers representing sides of a triangle: ";
cin >> side1 >> side2 >> side3;

//determine if it is a right triangle
sum = (side1 *side1) + (side2 * side2);
side3 = (side3 * side3);

if (sum == side3)
cout << "This is a right triangle.";
else
cout << "This is not a right triangle.";

return 0;

}
Now, this works fine on ( 0, 0, 0 ) and ( 3, 4, 5 ). But, on my machine, it
fails on ( 0.3, 0.4, 0.5 ). The reason is that you use == for variables of
type double which is not really what you want in cases where the internal
precission kicks in and screws the least significant digits of your
numbers.
Best

Kai-Uwe Bux

[M]

Fine, but won't our friend here learn more if we guide him (her?)
toward the answers rather than giving them to him? It just seems to
me that he won't learn very much if we do his homework for him.
On Sat, 10 Sep 2005 23:18:48 +1000, "John Carson"
<jc************ ****@netspace.n et.au> wrote:

"Oliver (Nospam)" <ok*****@arcor. de> wrote in message
news:43******* *************** @newsread2.arco r-online.net

I inserted corrections below:

"deanfamily 11" <de**********@v erizon.net> schrieb im Newsbeitrag
news:QKrUe.165$ R9.0@trnddc02.. .

#include <iomanip>
#include <iostream>

using namespace std;

int main()
{

//declare variables
int side1, side2, side3;
double sum;

//get sides from the user
cout << "Enter 3 integers representing sides of a triangle: ";
cin >> side1 >> side2 >> side3;

//determine if it is a right triangle
sum = (side1 ^ 2) + (side2 ^ 2);

"^" is taken from basic. In C you call the function "pow":
sum = pow(side1, 2) + pow(side2, 2);

side3 = (side3 ^ 2);

side3 = pow(side3, 2);

if (sum == side3)

Because floating numbers are never as precise, that equality can
always be guaranteed, even in case were equality is given, you should
write
if ( fabs(sum-side3) < 0.01 )

cout << "This is a right triangle.";
else
cout << "This is not a right triangle.";

return 0;

}

Given that the user is entering integers, you can test exactly (provided the
numbers are not too big) with:

sum = side1*side1 + side2*side2;
side3 = side3*side3;

if(sum == side3)
//

[Oliver \(Nospam\)]

Clever guy, hmm? Watching all around and pouring out very amusing comments
of - I guess - an elder veteran, watching for beginners to satisfy his own
needs of understatement. Very fine, Mr M anonymous! Very fine! Thanks a lot
for your good advices all around here. We love you very much. Thanks again
for visiting us.