Where is the wrong..?

[snapshooter]

salut..i am trying to make a program in order to calculate the y^2=a^2+b^2 and the code is:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

using namespace std;

double ypotinousa(double a, double b); /* function prototype*/

int main(int argc, char *argv[]) /*program begins*/

{
    
 double a,b,y,result;
 
 printf("Give the first number:\n");
 scanf("%d",&a);
 printf("Give the second number:\n");
 scanf("%d",&b);
 
 printf("The result is:%d\n",ypotinousa(a,b));
 
 system("PAUSE");	
 return 0;
 
}

double ypotinousa(double a, double b)
{
double y;

y=pow(a,2)+pow(b,2);

return sqrt(y);
}

This program is always return me zero after running..what is wrong? (i guess the function type..)

[snapshooter]

erase result in the code..is wrong

[Salem]

> using namespace std;
Get rid of this, this is C++
You're programming (and posting) C

> scanf("%d",&a);
Very bad - the conversion type and variable do not match - you're reading an int into a double
This does not work at all.

I suggest you grab the GNU compiler (one of these -> mingw, dev-c++, cygwin, djgpp)
Then you can do
gcc -W -Wall -ansi -pedantic -O2 hello.c
to spot these (and many others) errors

[Thantos]

also add in

printf("The result is:%d\n",ypotinousa(a,b));

to the type specifier problem mentioned by salem.

[snapshooter]

so what is the syntax for scanf in order to read double numbers..?

[Calavera]

scanf("%f", &a);

[Salem]

Or even scanf( "%lf", &mydouble );

Whilst "%f" is good for both floats and doubles when using printf(), the same is not true for scanf()

But using gcc with the compile options I posted previously would tell you this...