Question

I have absolutely no idea why it returns 2 for a=2 and b=2..

Any ideas?

#include <stdlib.h>

int main()
{
    double a,b,c;
    printf("a=");
    scanf("%d", &a);
    printf("b=");
    scanf("%d", &b);
    printf("c=");
    scanf("%d", &c);

    printf("x=%d", a+b);

    return 0;
}

Answer 1

The specifier "%d" expects an integer and you are passing the address of a double. Using the wrong specifiers in scanf leads to undefined behavior.

Also, using the wrong specifier in printf is the same thing. Because printf takes a variable number of arguments a + b which is a double can’t be converted to an integer.