接着问，请问怎么对地址做减法

[Viskag]

int width = 4*4；
byte* inPtr(该指针前面有48个byte，后面是16个byte)
......
__asm
{
mov esi, inPtr
mov eax, width
movdqu xmm1, [esi-eax]
}
原来的代码是movdqu xmm1，[esi+eax]，但是现在inPtr指向的源数据顺序反了，拷贝16byte后，要前移16，于是我就把esi+eax改成了esi-eax，报错error C2425: '-' : non-constant expression in 'second operand'，搜索了一下：
The complex addressing modes do not allow the subtraction of the index, its still a BASE address plus the INDEX then any displacement after as a constant. The real problem is there is no opcode to do this task.
这个看不懂，如果我想用减应该怎么写？

[Viskag]

厉害~

[zara]

将 eax 先取反下了：(寻址时，地址都是相加的)
__asm
{
mov esi, inPtr
mov eax, width
neg eax
movdqu xmm1, [esi+eax]
}