汇编:与32位寄存器进行64位乘法 [英] Assembly: 64 bit multiplication with 32-bit registers

问题描述

我对64位乘法的x86汇编中的实现有疑问.我已经尽我所能将代码发布了.我对其余的功能一无所知(很可能我在已经做的事情中犯了错误).任何方向将不胜感激.

I have a question regarding an implementation in x86 assembly of 64 bit multiplication. I've posted the code with as far as I was able to get in understanding it. I'm at a loss as to what the rest does (and it's possible I have made errors in what I've already done). Any direction would be appreciated.

dest at %ebp+8
x    at %ebp+12
y    at %ebp+16

movl        16(%ebp), %esi      //Move y into %esi
movl        12(%ebp), %eax      //Move x into %eax
movl        %eax, %edx          //Move x into %edx
sarl        $31, %edx            //Shift x right 31 bits (only sign bit remains)
movl        20(%ebp), %ecx      //Move the low order bits of y into %ecx
imull       %eax, %ecx          //Multiply the contents of %ecx (low order bits of y) by x
movl        %edx, %ebx          //Copy sign bit of x to ebx
imull       %esi, %ebx          //Multiply sign bit of x in ebx by high order bits of y
addl        %ebx, %ecx          //Add the signed upper order bits of y to the lower order bits (What happens when this overflows?)
mull        %esi                //Multiply the contents of eax (x) by y
leal        (%ecx,%edx), %edx           
movl        8(%ebp), %ecx
movl        %eax, (%ecx)
movl        %edx, 4(%ecx)

推荐答案

下面是64位乘法的算法:

Below is the algorithm of 64-bit multiplication:

x, y: 64-bit integer
x_h/x_l: higher/lower 32 bits of x
y_h/y_l: higher/lower 32 bits of y

x*y  = ((x_h*2^32 + x_l)*(y_h*2^32 + y_l)) mod 2^64
     = (x_h*y_h*2^64 + x_l*y_l + x_h*y_l*2^32 + x_l*y_h*2^32) mod 2^64
     = x_l*y_l + (x_h*y_l + x_l*y_h)*2^32

Now from the equation you can see that only 3(not 4) multiplication needed.

 movl 16(%ebp), %esi    ; get y_l
 movl 12(%ebp), %eax    ; get x_l
 movl %eax, %edx
 sarl $31, %edx         ; get x_h, (x >>a 31), higher 32 bits of sign-extension of x
 movl 20(%ebp), %ecx    ; get y_h
 imull %eax, %ecx       ; compute s: x_l*y_h
 movl %edx, %ebx
 imull %esi, %ebx       ; compute t: x_h*y_l
 addl %ebx, %ecx        ; compute s + t
 mull %esi              ; compute u: x_l*y_l
 leal (%ecx,%edx), %edx ; u_h += (s + t), result is u
 movl 8(%ebp), %ecx
 movl %eax, (%ecx)
 movl %edx, 4(%ecx)

您还可以检查此实现32位计算机上的64位算术

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