Why is %eax zeroed before a call to printf?

In the x86_64 ABI, if a function has variable arguments then AL (which is part of EAX) is expected to hold the number of vector registers used to hold arguments to that function.

In your example:

printf("%d", 1);

has an integer argument so there’s no need for a vector register, hence AL is set to 0.

On the other hand, if you change your example to:

printf("%f", 1.0f);

then the floating-point literal is stored in a vector register and, correspondingly, AL is set to 1:

movsd   LC1(%rip), %xmm0
leaq    LC0(%rip), %rdi
movl    $1, %eax
call    _printf

As expected:

printf("%f %f", 1.0f, 2.0f);

will cause the compiler to set AL to 2 since there are two floating-point arguments:

movsd   LC0(%rip), %xmm0
movapd  %xmm0, %xmm1
movsd   LC2(%rip), %xmm0
leaq    LC1(%rip), %rdi
movl    $2, %eax
call    _printf

---

As for your other questions:

puts is also zeroing out %eax right before the call though it only takes a single pointer. Why is this?

It shouldn’t. For instance:

#include <stdio.h>

void test(void) {
    puts("foo");
}

when compiled with gcc -c -O0 -S, outputs:

pushq   %rbp
movq    %rsp, %rbp
leaq    LC0(%rip), %rdi
call    _puts
leave
ret

and %eax is not zeroed out. However, if you remove #include <stdio.h> then the resulting assembly does zero out %eax right before calling puts():

pushq   %rbp
movq    %rsp, %rbp
leaq    LC0(%rip), %rdi
movl    $0, %eax
call    _puts
leave
ret

The reason is related to your second question:

This also happens before any call to my own void proc() function (even with -O2 set), but it is not zeroed when calling a void proc2(int param) function.

If the compiler doesn't see the declaration of a function then it makes no assumptions about its parameters, and the function could well accept variable arguments. The same applies if you specify an empty parameter list (which you shouldn’t, and it’s marked as an obsolescent C feature by ISO/IEC). Since the compiler doesn’t have enough information about the function parameters, it zeroes out %eax before calling the function because it might be the case that the function is defined as having variable arguments.

For example:

#include <stdio.h>

void function() {
    puts("foo");
}

void test(void) {
    function();
}

where function() has an empty parameter list, results in:

pushq   %rbp
movq    %rsp, %rbp
movl    $0, %eax
call    _function
leave
ret

However, if you follow the recommend practice of specifying void when the function accepts no parameters, such as:

#include <stdio.h>

void function(void) {
    puts("foo");
}

void test(void) {
    function();
}

then the compiler knows that function() doesn't accept arguments — in particular, it doesn’t accept variable arguments — and hence doesn’t clear %eax before calling that function:

pushq   %rbp
movq    %rsp, %rbp
call    _function
leave
ret

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.