64-assembly中的Imul

Question

I have this code:

mov rax, 0x93f3ffc2fbc7a1ce
mov rbx, 0x5862d8a05a385cbe
imul eax, ebx

How does imul work for 64-bit assembly? Will the overflow aex be written in the first 32 bits of rax?

Answer 1

Your code assembles to

0:  48 b8 ce a1 c7 fb c2    movabs rax,0x93f3ffc2fbc7a1ce
7:  ff f3 93
a:  48 bb be 5c 38 5a a0    movabs rbx,0x5862d8a05a385cbe
11: d8 62 58
14: 0f af c3                imul   eax,ebx

which uses the opcode 0F AF for imul . This instruction has 32-bit operand size so it only read EAX and EBX, and only writes EAX. This implicitly zero-extends into RAX , zeroing the upper 32 bits.

Unlike the 1-operand form of imul , the high-half of the 32x32 => 64-bit full multiply isn't written to EDX (or anywhere else like the high half of RAX); it's simply discarded or for efficiency not even calculated at all. See the documentation ; 2-operand imul reg, r/m32 is just like add reg, r/m32 or or reg, r/m32 - it doesn't do any special weird stuff.

Using mov rax, imm64 before this 32-bit multiply is completely pointless, mov eax,0xfbc7a1ce would give exactly identical results. (The imul doesn't destroy RBX, so the upper 32 bits of the value you put into RBX is still there if you want to read it later. It has no effect on the imul instruction, though.)

Even better, imul eax, ebx, 0xfbc7a1ce could have avoided a mov .