为什么 GCC 会发出“lea"?而不是“子"减法? [英] Why does GCC emit "lea" instead of "sub" for subtraction?

问题描述

我正在查看通过反汇编一些 C 程序生成的一些程序集，但我对我经常看到重复的单个优化感到困惑.

I am looking at some assembly that was generated by disassembling some C programs and I am confused by a single optimization that I see repeated frequently.

当我在 GCC 编译器上没有优化时使用 subl 指令进行减法，但是当我打开优化时(-O3 准确地说)编译器使用 leal 指令代替减法，示例如下:

When I have no optimizations on the GCC compiler uses the subl instruction for subtraction, but when I do have optimizations turned on (-O3 to be precise) the compiler uses a leal instruction instead of subtraction, example below:

没有优化:

83 e8 01     subl $0x1, %eax 

有优化

8d 6f ff     leal -0x1(%edi), %ebp 

这两条指令的长度都是 3 个字节，所以我在这里没有看到优化.有人可以帮我解释一下编译器的选择吗?

Both of these instructions are 3 bytes long, so I am not seeing an optimization here. Could someone help me out and try to explain the compiler's choice ?

任何帮助将不胜感激.

推荐答案

如果没有看到产生这个的原始 C 代码，很难判断.

It's hard to tell without seeing the original C code that produces this.

但如果我不得不猜测的话，那是因为 leal 允许在不破坏源寄存器的情况下不正确地进行减法.

But if I had to guess, it's because the leal allows the subtraction to be done out-of-place without destroying the source register.

这可以节省额外的寄存器移动.

This can save an extra register move.

第一个例子:

83 e8 01     subl $0x1, %eax 

覆盖 %eax 从而破坏原始值.

overwrites %eax thereby destroying the original value.

第二个例子:

8d 6f ff     leal -0x1(%edi), %ebp 

将 %edi - 1 存储到 %ebp 中.%edi 保留以备将来使用.

stores %edi - 1 into %ebp. %edi is preserved for future use.

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