Printf with no arguments explanation

Question

I understand that if printf is given no arguments it outputs an unexpected value.

Example:

#include <stdio.h>

int main() {
    int test = 4 * 4

    printf("The answer is: %d\n");
    return 0;
}

This returns a random number. After playing around with different formats such as %p, %x etc, it doesn't print 16(because I didn't add the variable to the argument section) What i'd like to know is, where are these values being taken from? Is it the top of the stack? It's not a new value every time I compile, which is weird, it's like a fixed value.

Answer 1

printf("The answer is: %d\n");

invokes undefined behavior. C requires a conversion specifier to have an associated argument. While it is undefined behavior and anything can happen, on most systems you end up dumping the stack. It's the kind of trick used in format string attacks.

Answer 2

It is called undefined behavior and it is scary (see this answer).

If you want an explanation, you need to dive into implementation specific details. So study the generated source code (e.g. compile with gcc -Wall -Wextra -fverbose-asm + your optimization flags, then look into the generated .s assembly file) and the ABI of your system.