问题描述
我正在研究《计算机系统》，《程序员的视角》 （第3版），《实践问题3.3》包含以下内容：
I'm working through Computer Systems, A Programmer's Perspective (3rd edition), and Practice Problem 3.3 contains the following line:
movb $0xF, (%ebx)
我应该找出这行x86-64汇编出了什么问题，答案键指出：不能将％ebx用作地址寄存器，这对我来说没有意义。我的理解是，该行打算将0xF复制到主内存中的某个位置，但是％ebx是32位寄存器，在64位计算机上内存地址为64位宽，因此％ebx无法保存内存地址，因此它不能被取消引用（取消引用是％ebx周围的括号表示的，对吗？）。但是，回头看书中的几页（如果有的话，请参阅第183页），有一个示例详细说明了五个mov操作数-目标组合，其中之一是：
I'm supposed to find out what's wrong with this line of x86-64 assembly, and the answer key states: "Cannot use %ebx as address register", which doesn't make sense to me. My understanding is that this line intends to copy 0xF to a location in main memory, however %ebx is a 32-bit register, memory addresses are 64 bits wide on 64-bit machines, and so %ebx cannot hold a memory address, therefore it cannot be dereferenced (dereferencing is what the parentheses around %ebx represent, correct?). However, looking a few pages back in the book (page 183, if you have it) there is an example detailing the five mov operand--destination combinations, one of which is:
movb $-17, (%esp) Immediate--Memory, 1 byte
％esp是32位寄存器，就像％ebx一样！此示例显示了将字节值移至已取消引用的32位寄存器！这对我来说没有任何意义，因为％esp如何包含64位地址？我会完全误解汇编吗？
%esp is a 32-bit register just like %ebx! And this example shows a byte value being moved to a dereferenced 32-bit register! Which doesn't make sense to me, because how can %esp contain a 64-bit address? Do I completely misunderstand assembly?
推荐答案
您是对的，
movb $-17, (%esp) Immediate--Memory, 1 byte
不应被允许。实际上，作者将其发布为错字。
查看他们的（Ctrl-F表示 p 。183）。
should not be allowed. In fact the authors have posted this as a typo.Check out their errata list (Ctrl-F for "p. 183").
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