当您使用内存覆盖前缀但所有操作数均为寄存器时会发生什么? [英] What happens when you use a memory override prefix but all the operands are registers?

问题描述

当您使用内存覆盖前缀但所有操作数都是寄存器时会发生什么?

What happens when you use a memory override prefix but all the operands are registers?

因此，假设您编写的代码为mov eax, ebx或add eax, ebx，默认值为32位，但使用了67h覆盖.

So, let's say you code mov eax, ebx or add eax, ebxand the default is 32-bit but you use a 67h override.

处理器如何处理这种情况?

How does the processor handle that situation?

推荐答案

《英特尔软件开发人员手册*》第2卷第2.1节详细介绍了每个指令前缀的行为.它说，将地址大小前缀(67h)与没有内存操作数的指令一起使用是保留的，并且可能导致不可预测的行为.

The Intel Software Developer's Manual*, volume 2, section 2.1, details the behavior of each instruction prefix. It says use of the address-size prefix (67h) with an instruction that doesn't have a memory operand is reserved and may cause unpredictable behavior.

操作数大小前缀(66h)可用于在16位和32位操作数大小之间切换，也可作为某些SSE2/SSE3/SSSE3/SSE4指令的强制前缀.保留其他使用权，并可能导致无法预料的行为.

The operand-size prefix (66h) may be used to switch between 16- and 32-bit operand sizes and also as a mandatory prefix with certain SSE2/SSE3/SSSE3/SSE4 instructions. Other use is reserved and may cause unpredictable behavior.

段覆盖前缀与任何分支指令一起保留.

The segment override prefixes are reserved with any branch instruction.

* https://software.intel.com/zh-cn/articles /intel-sdm

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