为何％eax中一个归零调用printf之前？ [英] Why is %eax zeroed before a call to printf?

问题描述

我试图拿起一点点的x86。我编译在64位MAC用gcc -S -O0。

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

code在C：

printf("%d", 1);

输出：

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

我不明白为什么'的printf'调用之前％EAX清零。由于的printf 返回打印到％EAX 中的字符数我最好的猜测是归零，以prepare它为的printf ，但我会假设的printf 必须负责得到它准备好。此外，相反，如果我把我自己的函数 INT testproc（INT P1）， GCC 认为没有必要prepare ％EAX 。所以，我不知道为什么 GCC 治疗的printf 和 testproc 不同。

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.

推荐答案

从 x86_64的ABI ：

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

printf的是具有可变参数的函数，以及用于矢量寄存器的数目是零。

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

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