如何BSWAP 64位寄存器的低32位? [英] How to BSWAP the lower 32-bit of 64-bit register?

问题描述

我一直在寻找有关如何将 BSWAP 用于 64 位寄存器的低 32 位子寄存器的答案.例如，0x0123456789abcdef 在 RAX 寄存器内，我想用一条指令将其更改为 0x01234567efcdab89(出于性能考虑).

I've been looking for the answer for how to use BSWAP for lower 32-bit sub-register of 64-bit register. For example, 0x0123456789abcdef is inside RAX register, and I want to change it to 0x01234567efcdab89 with a single instruction (because of performance).

所以我尝试了以下内联函数:

So I tried following inline function:

#define BSWAP(T) {  
    __asm__ __volatile__ (  
            "bswap %k0" 
            : "=q" (T)  
            : "q" (T)); 
}

结果是0x00000000efcdab89.我不明白为什么编译器会这样.有人知道有效的解决方案吗?

And the result was 0x00000000efcdab89. I don't understand why the compiler acts like this. Does anybody know the efficient solution?

推荐答案

啊，是的，我现在明白问题所在了:

Ah, yes, I understand the problem now:

x86-64 处理器隐式零扩展 在执行 32 位操作(在 %eax、%ebx 等上)时，32 位寄存器会注册为 64 位.据我了解，这是为了保持与旧代码的兼容性，这些代码需要这些寄存器的 32 位语义.

the x86-64 processors implicitly zero-extend the 32-bit registers to 64-bit when doing 32-bit operations (on %eax, %ebx, etc). This is to maintain compatibility with legacy code that expects 32-bit semantics for these registers, as I understand it.

所以恐怕没有办法仅对 64 位寄存器的低 32 位执行 ror.你必须使用一系列的指令......

So I'm afraid that there is no way to do ror on just the lower 32 bits of a 64-bit register. You'll have to do use a series of several instructions...

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