“ movl”组装中的说明
我正在学习组装和阅读“计算机系统：程序员的观点”。实际上，问题3.3，它说movl％eax，％rdx将产生错误。答案键说movl％eax，％dx目标操作数不正确的大小。我不确定这是否是错别字，但是我的问题是：movl％eax，％rdx法律指令吗？我认为它正在以％eax中的32位移动零扩展名为％rdx，这不会以> movzql生成
一项指令生成带有寄存器的4字节值，因为目的地将填充上4个字节的零（来自书籍）。
我尝试编写一些C代码来生成它，但是我总是会得到movslq％eax，％rdx（GCC 4.8.5 -og）。我完全困惑。
I am learning assembly and reading "Computer Systems: A programmer's perspective". In Practice Problem 3.3, it says movl %eax,%rdx will generate an error. The answer keys says movl %eax,%dx Destination operand incorrect size. I am not sure if this is a typo or not, but my question is: is movl %eax,%rdx a legal instruction? I think it is moving the 32 bits in %eax with zero extension to %rdx, which will not be generated as movzql since
an instruction generating a 4-byte value with a register as the destination will fill the upper 4 bytes with zeros` (from the book).
I tried to write some C code to generate it, but I always get movslq %eax, %rdx(GCC 4.8.5 -Og). I am completely confused.
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
GNU汇编程序不接受
movl％eax，％rdx。对于编码而言，这也没有意义，因为mov必须具有单个操作数大小（如果需要的话，请使用前缀字节），而不是两个不同尺寸的操作数。您想要的效果是通过
movl％eax，％edx来实现的，因为将写入32位寄存器的写入始终为零 - 零 - 在相应的64位寄存器中。请参阅为什么在32位寄存器上x86-64说明完整的上部64位注册？。movzlq％eax，％rdx可能是合乎逻辑的，但是不支持它，因为它是多余的。The GNU assembler doesn't accept
movl %eax,%rdx. It also doesn't make sense for the encoding, sincemovmust have a single operand size (using a prefix byte if needed), not two different sized operands.The effect you want is achieved by
movl %eax, %edxsince writes to a 32-bit register always zero-extend into the corresponding 64-bit register. See Why do x86-64 instructions on 32-bit registers zero the upper part of the full 64-bit register?.movzlq %eax, %rdxmight make logical sense, but it's not supported since it would be redundant.