Question

I am trying to pick up a little x86. I am compiling on a 64bit mac with gcc -S -O0.

Code in C:

printf("%d", 1);

Output:

movl    $1, %esi
leaq    LC0(%rip), %rdi
movl    $0, %eax        ; WHY?
call    _printf

I do not understand why %eax is cleared to 0 before ‘printf’ is called. Since printf returns the number of characters printed to %eax my best guess it is zeroed out to prepare it for printf but I would have assumed that printf would have to be responsible for getting it ready. Also, in contrast, if I call my own function int testproc(int p1), gcc sees no need to prepare %eax. So I wonder why gcc treats printf and testproc differently.

Answer 1

From the x86_64 System V ABI register usage table:

• %rax       temporary register; with variable arguments
  passes information about the number of vector
  registers used; 1st return register
  …

printf is a function with variable arguments, and the number of vector registers used is zero.

Note that printf must check only %al, because the caller is allowed to leave garbage in the higher bytes of %rax. (Still, xor %eax,%eax is the most efficient way to zero %al)

See the this Q&A and the x86 tag wiki for more details, or for up-to-date ABI links if the above link is stale.