Question

您如何知道如何在以下汇编代码中填写下划线：

mov_ %eax, (%rsp)

字节为“b”，字为“w”，双字为“l”，四字为“q”。寄存器的语法应该（我认为）表明移动了多少数据。我查看了我的书，似乎无法确定如何确定。

有没有通用的方法来解决这个问题？

Answer 1

If one operand of mov is a register, then the size is implied: AL is one byte, AX two, EAX four and RAX eight. If one operand is a memory location and the other an immediate value, you have to specify the desired size:

mov  BYTE PTR [RAX], 1    ;  *(uint8_t *)(rax) = 1
mov  WORD PTR [RAX], 1    ; *(uint16_t *)(rax) = 1
mov DWORD PTR [RAX], 1    ; *(uint32_t *)(rax) = 1
mov QWORD PTR [RAX], 1    ; *(uint64_t *)(rax) = 1

The above is Intel syntax as used by the architecture manual. In AT&T syntax, the first line would be movb $1, (%rax); AT&T syntax uses suffixed instructions like movb, movw, movd and movq to indicate operand widths.

Answer 2

AT&T assembler is not ambiguous, so it must match the operands, even if the operands(registers) contain enough size information.

%eax is a 32-bit register (just like %al is 8-bit, %ax is 16-bit and %rax is 64-bit).

While %rsp in (%rsp) is 64-bit, the expression (%rsp) is a move to the memory location held in %rsp, and as such has no register size associated to it.

So the mov(e) is 32-bit, and the suffix is l.

如何判断x86中寄存器的大小？

2 个答案: