Question

This code compiles fine, but when I try to run it, it fails.

int main(int argc, char** argv) {
    int c = 4;
    int b = 5;
    int a = a & b;
    printf(a);

    return (EXIT_SUCCESS);
}

What am I doing wrong?

Answer 1

The first argument to printf is the format string, a char*. Read this for more information about constructing the format string.

printf(a);

should be:

printf("%d", a);

Otherwise a will be treated as the address of the start of a string, and cause undefined behaviour when it’s dereferenced (there’s a good chance it’ll crash).

Also, I’m assuming that the above code isn’t the exact code you’re talking about, since it won’t compile as it is.