问题
This is a rather naive question. I'm just playing around with allocating some executable memory and manually assembling some x86 code to run in it. I'm a bit confused by how addressing works in 64-bit mode. If I have a variable in my C code, and I want to move the contents of RAX into this variable, which form of the MOV instruction should I use? (This isn't using inline assembly, so I can't get the compiler to do it for me; I just have the value of &var to play with.)
回答1:
Using Intel syntax, it will be something like mov [var],rax, where var is basically the 64-bit immediate address.
I think this what Intel call MOV moffs64*,RAX in "Intel® 64 and IA-32 Architectures
Software Developer’s Manual".
来源：https://stackoverflow.com/questions/4318816/moving-contents-of-rax-to-c-variable-x86-64-asm