64位汇编，何时使用较小尺寸的寄存器 [英] 64 bit assembly, when to use smaller size registers

问题描述

我知道在x86_64汇编中有(64位)rax寄存器，但也可以作为32位寄存器eax，16位，ax和8位等来访问它.在什么情况下我不仅会使用完整的64位，为什么，会有什么优势?

I understand in x86_64 assembly there is for example the (64 bit) rax register, but it can also be accessed as a 32 bit register, eax, 16 bit, ax, and 8 bit, al. In what situation would I not just use the full 64 bits, and why, what advantage would there be?

作为示例，使用这个简单的hello world程序:

As an example, with this simple hello world program:

section .data
msg: db "Hello World!", 0x0a, 0x00
len: equ $-msg

section .text
global start

start:
mov rax, 0x2000004      ; System call write = 4
mov rdi, 1              ; Write to standard out = 1
mov rsi, msg            ; The address of hello_world string
mov rdx, len            ; The size to write
syscall                 ; Invoke the kernel
mov rax, 0x2000001      ; System call number for exit = 1
mov rdi, 0              ; Exit success = 0
syscall                 ; Invoke the kernel

rdi和rdx至少只需要8位而不是64位，对吗?但是，如果我分别将它们更改为dil和dl(它们的低8位等效项)，则该程序将进行汇编和链接，但不会输出任何内容.

rdi and rdx, at least, only need 8 bits and not 64, right? But if I change them to dil and dl, respectively (their lower 8-bit equivalents), the program assembles and links but doesn't output anything.

但是，如果我使用eax，edi和edx，它仍然可以工作，那么我应该使用那些而不是完整的64位吗?为什么或为什么不呢?

However, it still works if I use eax, edi and edx, so should I use those rather than the full 64-bits? Why or why not?

推荐答案

首先要从内存中加载较小的值(例如8位)(读取char，处理数据结构，反序列化网络数据包) ，等等.)进入寄存器.

First and foremost would be when loading a smaller (e.g. 8-bit) value from memory (reading a char, working on a data structure, deserialising a network packet, etc.) into a register.

MOV AL, [0x1234]

与

MOV RAX, [0x1234]
SHR RAX, 56
# assuming there are actually 8 accessible bytes at 0x1234,
# and they're the right endianness; otherwise you'd need
# AND RAX, 0xFF or similar...

或者，当然，将所述值写回内存中.

Or, of course, writing said value back to memory.

(如6年后一样进行编辑):

由于这种情况不断出现:

Since this keeps coming up:

MOV AL, [0x1234]

• 仅读取0x1234的单个字节的内存(反之只会覆盖单个字节的内存)
• 保留RAX其他56位中的任何内容
  • 这会在RAX的过去值和将来值之间建立依赖关系，因此CPU无法使用
    • only reads a single byte of memory at 0x1234 (the inverse would only overwrite a single byte of memory)
    • keeps whatever was in the other 56 bits of RAX
      • This creates a dependency between the past and future values of RAX, so the CPU can't optimise the instruction using register renaming.

      对比:

      MOV RAX, [0x1234]
      

      • 从0x1234开始读取8个字节的内存(反之会覆盖8个字节的内存)
      • 覆盖RAX的全部
      • 假定内存中的字节与CPU具有相同的字节序(在网络数据包中通常不正确，因此几年前我的SHR指令)
      • reads 8 bytes of memory starting at 0x1234 (the inverse would overwrite 8 bytes of memory)
      • overwrites all of RAX
      • assumes the bytes in memory have the same endianness as the CPU (often not true in network packets, hence my SHR instruction years ago)

      还要注意的重要事项:

      MOV EAX, [0x1234]
      

      • 从0x1234开始读取4个字节的内存(反之会覆盖4个字节的内存)
      • 覆盖所有RAX ，但高位将全部为零
        • 请参阅:为什么大多数x64指令将32位寄存器的高位归零
        • reads 4 bytes of memory starting at 0x1234 (the inverse would overwrite 4 bytes of memory)
        • overwrites all of RAX, but the high bits will all be zero
          • see: Why do most x64 instructions zero the upper part of a 32 bit register

          然后，如评论中所述，有:

          Then, as mentioned in the comments, there is:

          MOVZX EAX, byte [0x1234]
          

          • 仅读取0x1234的单个字节的内存
          • 扩展该值，以用零填充所有EAX(并因此填充RAX)(消除依赖性并允许寄存器重命名优化).

          在所有这些情况下，如果要从'A'寄存器中写入到内存中，则必须选择宽度:

          In all of these cases, if you want to write from the 'A' register into memory you'd have to pick your width:

          MOV [0x1234], AL   ; write a byte (8 bits)
          MOV [0x1234], AX   ; write a word (16 bits)
          MOV [0x1234], EAX  ; write a dword (32 bits)
          MOV [0x1234], RAX  ; write a qword (64 bits)
          

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