Print value of function without a variable in C-CodePudding

let's assume the following code in c:

#include <stdio.h>
#include <cs50.h>

int test (int a, int b);

int main(void)
{
   test(2,3);
}

int test (int a, int b)
{
 int c = a b;
 printf("%d \n", test(a,b));
 return c;

}

why is it not possible to print the value of test without having to save it in a variable before and print the variable? I get the error:

function.c:12:1: error: all paths through this function will call itself [-Werror,-Winfinite-recursion]

Thank you!

#include <stdio.h>
#include <cs50.h>

int test (int a, int b);

int main(void)
{
   test(2,3);
}

int test (int a, int b)
{
 int c = a b;
 printf("%d \n", test(a,b));
 return c;

}

CodePudding user response：

The error message is quite clear. The test-function calls itself. And inside that call, it calls itself again, (and again, and again...).

It will never complete.

This is a type of infinite loop, commonly called infinite recursion.

Perhaps what you want is?

#include <stdio.h>
#include <cs50.h>

int test (int a, int b);

int main(void)
{
   test(2,3);
}

int test (int a, int b)
{
 int c = a b;
 printf("%d \n", c); // Show the result of the calculation
                     // but without calling this function again.
 return c;

}

CodePudding user response：

As the compiler message says, the function will call itself, because, in printf("%d \n", test(a,b));, the code test(a,b) calls test. Inside that call to test, the function will call itself again, and this will repeat forever (up to the limits of the C implementation).

To print the return value of the function, do it outside the function:

#include <stdio.h>

int test(int a, int b);

int main(void)
{
    printf("%d\n", test(2, 3));
}

int test(int a, int b)
{
    return a b;
}