imul 汇编指令 - 一个操作数？
我正在使用运行时调试器。
EAX：0000 0023 EDX：5555 5556
imul edx
EAX：aaaa aac2 EDX: 0000 000b
我完全困惑了，无法弄清楚这个乘法是如何工作的。这里发生了什么事？我注意到在一个类似的问题中imul ebx ;结果是 EDX:EAX 但我不明白 EDX:EAX 表示法：/
I am using a run-time debugger.
EAX: 0000 0023
EDX: 5555 5556
imul edx
EAX: aaaa aac2
EDX: 0000 000b
I am utterly confused, and can't figure out how this multiply is working. What's happening here? I notice in a similar question here that imul ebx ; result in EDX:EAX I don't understand the EDX:EAX notation though :/
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当
imul的单操作数形式传递一个 32 位参数时，它实际上意味着EAX * src，其中EAX和源操作数都是32 位寄存器或内存。两个 32 位值的乘积不一定适合 32 位：完整的乘法结果最多可以占用 64 位。答案的高32位将写入
EDX寄存器，低32位写入EAX寄存器；这用EDX:EAX表示法表示。在使用
imul edx的情况下，您将得到EDX:EAX = EAX * EDX。显式源操作数可以成为隐式操作数之一，甚至可以将EAX平方为EDX:EAX。如果您只需要结果的低 32 位，请使用
imul的 2 操作数形式；它运行速度更快并且没有任何隐式操作数（因此您可以使用最方便的任何寄存器）。imul ecx, esi会按照您的预期执行ecx *= esi操作，而不会触及EAX或EDX。就像 C 中unsigned x=...;x *= y;结果的宽度与输入的宽度相同。imul也有直接形式：imul ecx, ebx, 1234的作用是ecx = ebx * 1234。许多汇编器将接受imul ecx, 1234作为imul ecx, ecx, 1234的简写。这些 32x32 => 32 位形式的
imul对于有符号或无符号都能正常工作；单操作数mul和imul的结果仅在上半部分（在EDX中）不同，而不是下半部分EAX输出。请参阅 Intel 的
imul指令参考手册条目。When the one-operand form of
imulis passed a 32 bit argument, it effectively meansEAX * srcwhere bothEAXand the source operand are 32-bit registers or memory.The product of two 32 bit values doesn't necessarily fit in 32 bits: the full multiply result can take up to 64 bits. The high 32 bits of the answer will be written to the
EDXregister and the low 32 bits to theEAXregister; this is represented with theEDX:EAXnotation.In your case with
imul edx, you getEDX:EAX = EAX * EDX. It's fine for the explicit source operand to be one of the implicit operands, evenEAXto square intoEDX:EAX.If you only want the low 32 bits of the result, use the 2-operand form of
imul; it runs faster and doesn't have any implicit operands (so you can use whatever registers are most convenient).imul ecx, esidoesecx *= esilike you'd expect, without touchingEAXorEDX. It's like C whereunsigned x=...;x *= y;has the same width for the result as the inputs.imulalso has an immediate form:imul ecx, ebx, 1234doesecx = ebx * 1234. Many assemblers will acceptimul ecx, 1234as short-hand forimul ecx, ecx, 1234.These 32x32 => 32-bit forms of
imulwork correctly for signed or unsigned; the results of one-operandmulandimulonly differ in the upper half (inEDX), not the low-halfEAXoutput.See Intel's instruction reference manual entry for
imul.