c - 为什么 gcc 使用 movl 而不是 push 来传递函数参数？

Question

注意这段代码:

#include <stdio.h>
void a(int a, int b, int c)
{
    char buffer1[5];
    char buffer2[10];
}

int main()
{
    a(1,2,3); 
}

在那之后:

gcc -S a.c

该命令以汇编形式显示我们的源代码。

现在我们可以在主函数中看到，我们从来没有使用“push”命令来推送参数a函数入栈。它使用“移动”而不是那个

main:
 pushl %ebp
 movl %esp, %ebp
 andl $-16, %esp
 subl $16, %esp
 movl $3, 8(%esp)
 movl $2, 4(%esp)
 movl $1, (%esp)
 call a
 leave

为什么会这样？它们有什么区别？

Answer 1

这是the gcc manual不得不说:

-mpush-args
-mno-push-args
    Use PUSH operations to store outgoing parameters. This method is shorter and usually
    equally fast as method using SUB/MOV operations and is enabled by default. 
    In some cases disabling it may improve performance because of improved scheduling
    and reduced dependencies.

 -maccumulate-outgoing-args
    If enabled, the maximum amount of space required for outgoing arguments will be
    computed in the function prologue. This is faster on most modern CPUs because of
    reduced dependencies, improved scheduling and reduced stack usage when preferred
    stack boundary is not equal to 2. The drawback is a notable increase in code size.
    This switch implies -mno-push-args. 

显然 -maccumulate-outgoing-args 默认启用，覆盖 -mpush-args。使用 -mno-accumulate-outgoing-args 显式编译会恢复到 PUSH 方法，此处。

---

2019 年更新:自奔腾 M 以来，现代 CPU 已经拥有高效的 p​​ush/pop。
-mno-accumulate-outgoing-args(并使用推送)最终在 2014 年 1 月成为 -mtune=generic 的默认值。