Question

Here is my program:

#include <stdio.h>
int main()
{
    int a=0x09;
    int b=0x10;
    unsigned long long c=0x123456;
    printf("%x %llx\n",a,b,c);//in "%llx", l is lowercase of 'L', not digit 1
    return 0;
}

the output was:

9 12345600000010

I want to know:

1. how function printf() is executed?
2. what will happen if the number of arguments isn’t equal to that of formats?

please help me and use this program as an example to make an explanation.

Answer 1

The problem is that your types don’t match. This is undefined behavior.

Your second argument b does not match the type of the format. So what’s happening is that printf() is reading past the 4 bytes holding b (printf is expecting an 8-byte operand, but b is only 4 bytes). Therefore you’re getting junk. The 3rd argument isn’t printed at all since your printf() only has 2 format codes.

Since the arguments are usually passed consecutively (and adjacent) in memory, the 4 extra bytes that printf() is reading are actually the lower 4 bytes of c.

So in the end, the second number that’s being printed is equal to b + ((c & 0xffffffff) << 32).

But I want to reiterate: this behavior is undefined. It’s just that most systems today behave like this.