只有一个操作数的 SHR [英] SHR with only one operand

问题描述

我正在研究一些 32 位 x86 汇编代码分析项目.现在我真的对以下说明感到困惑:

I am working on some 32-bit x86 assembly code analysis projects. Right now I am really confused with the following instruction:

shr %edx

我理解操作码shr​​的语义，据我所知，这个操作码需要两个操作数.那么，谁能给我一些有关上述说明含义的指南?是否等于以下指令(在 AT&T 语法中)?

I understand the semantics of opcode shr, and as far as I can see, this opcode needs two operands. So, could anyone give me some guide on the meaning of the above instruction? Is it equal to the following instruction (in AT&T syntax)?

shr $2, %edx

推荐答案

正如任何操作码参考都会告诉您的那样，只有一个操作的 SHR 意味着您只移动一位.

As any opcode reference will tell you, SHR with only one operation implies that you are shifting only one bit.

您可能想要找到在线说明参考或获取英特尔指令参考.

You may want to either find an online instruction reference or grab a PDF copy of the Intel instruction references.

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