为什么 GCC 生成 mov %eax,%eax 是什么意思? [英] Why did GCC generate mov %eax,%eax and what does it mean?

问题描述

GCC 4.4.3 生成了以下 x86_64 程序集.令我困惑的部分是 mov %eax,%eax.将寄存器移至自身?为什么?

GCC 4.4.3 generated the following x86_64 assembly. The part that confuses me is the mov %eax,%eax. Move the register to itself? Why?

   23b6c:       31 c9                   xor    %ecx,%ecx        ; the 0 value for shift
   23b6e:       80 7f 60 00             cmpb   $0x0,0x60(%rdi)  ; is it shifted?
   23b72:       74 03                   je     23b77
   23b74:       8b 4f 64                mov    0x64(%rdi),%ecx  ; is shifted so load shift value to ecx
   23b77:       48 8b 57 38             mov    0x38(%rdi),%rdx  ; map base
   23b7b:       48 03 57 58             add    0x58(%rdi),%rdx  ; plus offset to value
   23b7f:       8b 02                   mov    (%rdx),%eax      ; load map_used value to eax
   23b81:       89 c0                   mov    %eax,%eax        ; then what the heck is this? promotion from uint32 to 64-bit size_t?
   23b83:       48 d3 e0                shl    %cl,%rax         ; shift rax/eax by cl/ecx
   23b86:       c3                      retq   

这个函数的C++代码是:

The C++ code for this function is:

    uint32_t shift = used_is_shifted ? shift_ : 0;
    le_uint32_t le_map_used = *used_p();
    size_t map_used = le_map_used;
    return map_used << shift;

le_uint32_t 是一个在 big-endian 机器上包装字节交换操作的类.在 x86 上它什么也不做.used_p() 函数根据映射基址 + 偏移量计算一个指针，并返回一个正确类型的指针.

An le_uint32_t is a class which wraps byte-swap operations on big-endian machines. On x86 it does nothing. The used_p() function computes a pointer from the map base + offset and returns a pointer of the correct type.

推荐答案

在 x86-64 中，32 位指令隐式零扩展:位 32-63 被清除 (以避免错误依赖).所以有时这就是为什么你会看到看起来很奇怪的说明.(是 mov %esi, %esi在 x86-64 上是否无操作?)

In x86-64, 32-bit instructions implicitly zero-extend: bits 32-63 are cleared (to avoid false dependencies). So sometimes that's why you'll see odd-looking instructions. (Is mov %esi, %esi a no-op or not on x86-64?)

然而，在这种情况下，前面的 mov-load 也是 32 位的，所以 %rax 的高半部分已经被清除了.mov %eax, %eax 似乎是多余的，显然只是 GCC 错过了优化.

However, in this case the previous mov-load is also 32-bit so the high half of %rax is already cleared. The mov %eax, %eax appears to be redundant, apparently just a GCC missed optimization.

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