MOVSX汇编指令如何工作? [英] How does MOVSX assembly instruction work?

问题描述

在以下示例中，汇编指令MOVSX如何工作:

How does the assembly instruction MOVSX work in this following example:

MOVSX ECX,BYTE PTR DS:[EDX]

在这种情况下，这是寄存器的状态:

In this case, here are the state of the registers:

ECX = 0000000F   
EDX = 0012FD9F 

根据我的想法，它需要[EDX] = 9F的最后一个字节，将其移至ECX，然后对其进行符号扩展以匹配16位= 0000009F.但是，实际结果是00000016.有人可以帮我解释我错了吗?

From what I thought, it takes last bytes of [EDX] = 9F, moves it to ECX and then sign extends it to match 16 bits = 0000009F. However, the actual result is 00000016. Can someone help explain where I'm wrong?

推荐答案

部分正确.但是:

BYTE PTR DS:[EDX]获取位于EDX中保存的地址中的字节.该字节被复制到ECX的最低有效字节中，其余字节用该字节的符号填充.

BYTE PTR DS:[EDX] obtains the byte located at the address held in EDX. This byte is copied to ECX into the least significant byte and the rest is filled with the sign of the byte.

对于意外的结果，这意味着在内存地址 ¹ 0x12FD9F中位于字节0x16.

For your unexpected result, this means that at the memory address¹ 0x12FD9F the byte 0x16 is located.

注意:

• 段覆盖前缀DS:在这里不是必需的. [EDX]自动引用DS.

• the Segment Override Prefix DS: isn't necessary here. [EDX] automatically refers to DS.

¹ 内存地址"在这里是指虚拟或物理内存

¹ "memory address" refers to either virtual or physical memory here

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