不带后缀的汇编mov指令 [英] Assembly mov instruction without suffix
问题描述
我有以下来自反汇编程序的mov指令(不带后缀).
I have the following mov instruction (without the suffix) from a disassembler.
mov %dx,(%eax)
指令后缀是什么?首先，我认为目标寄存器确定了后缀，但是根据我正在阅读的书，我猜它是由最小"寄存器确定的.因此，在这种情况下
What would be the instruction suffix? First I thought that the destination register determines the suffix, however according to the book I'm reading, I guess it's determined by the "smallest" register. So in this case would be
movw %dx, (%eax)
因为％dx(16位字寄存器)是最小的. 我的推理正确吗? (有时CSAPP的书有些混乱，没有清楚地说明细节.)
since %dx (16-bit word register) is the smallest one. Is my reasoning correct? (Sometimes the CSAPP book is a little bit confusing, doesn't explain details clearly).
推荐答案
目标不是示例中的寄存器，而是源，是寄存器.因此操作数大小为16位，因此AT& T将使用movw.
The destination isn't a register in your examples, it's the source that's a register. So the operand-size is 16-bit, thus AT&T would use movw.
目标是内存中的2个字节，由32位寻址模式选择. mov要求源和目标必须具有相同的宽度.如果至少一个操作数是一个寄存器，那么它将唯一地确定操作数的大小.
The destination is 2 bytes in memory, selected by a 32-bit addressing mode. mov requires both source and dest to be the same width. If at least one operand is a register, that uniquely determines the operand size.
对于mov $123, (%eax)之类的内容，您需要一个明确的后缀，因为这两个操作数都不是寄存器.
You need an explicit suffix for something like mov $123, (%eax) because neither operand is a register.
您最小"的想法完全是虚假的. movl %eax, (%bx)之所以是movl，是因为寄存器操作数是32位，并且通过16位寻址模式选择了4字节的目的地.
Your idea of "smallest" is totally bogus. movl %eax, (%bx) is movl because the register operand is 32 bits, and the 4-byte destination is selected by a 16-bit addressing mode.
一个或多个处于寻址模式的寄存器对操作数大小的影响为零.地址大小和操作数大小是独立的，您可以覆盖其中一个，但不能覆盖另一个. (这就是为什么操作数大小(0x66)和地址大小(0x67)有单独的机器代码前缀字节的原因.
The register or registers in the addressing mode have zero effect on the operand-size. Address-size and operand-size are independent, and you can override one but not the other. (That's why there are separate machine-code prefix bytes for operand-size (0x66) and address-size (0x67).
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