x 86彙編語言：移位乘以64位答案

Question

NOTICE: This question does relate to my homework/classwork. The textbook is poorly written so I can't really rely on it. 

我正在使用linux x86 x86彙編語言，並試圖找出如何使用shift操作數將兩個32位數字相乘。 我還需要找到一種方法將64位答案存儲到兩個單獨的寄存器中，因爲每個寄存器只有32位。我知道，向左移一次相當於乘以二，向右移兩分，但這是我現在確定的。任何幫助將不勝感激，解釋不僅僅是答案。

Answer 1

我想這應該這樣做：

mult: 
     #save all caller-saved regs 
     #move arg1 to %edi, arg2 to %esi 

     xorl %eax, %eax   #\ 
     xorl %edx, %edx   #--clear a 64-bit accumulator 
     movl $31, %ecx   #set up shift-count 

.L1: movl %edi, %ebx   #copy of arg1 
     xorl %ebp, %ebp   #zero scratch-register 
     shll %cl, %ebx   #isolate bit from arg1 
     sarl $31, %ebx   #and turn into mask 
     andl %esi, %ebx   #AND arg2 with bitmask 
     xorl $31, %ecx   #invert shift-count 
     shldl %cl, %ebx, %ebp #shift upper bits into scratch-reg 
     shll %cl, %ebx   #adjust lower bits 
     addl %ebx, %eax   #\ 
     addl %ebp, %edx   #-- accumulate results 
     xorl $31, %ecx   #restore shift-count 
     decl %ecx    #change shift to next bit 
     jno .L1      #if ecx == -1, done! 

     #restore caller-saved regs 
     #done, return value in edx:eax 

注意這把參數爲無符號。