x86-64递归乘法函数

Question

我试图在x86-64程序集中创建一个函数，它可以将两个数字x和y相乘，并在不使用循环或mul操作符的情况下返回输出。这是我想出的代码，但我不知道它为什么总是返回y。任何帮助都是非常感谢的。X在edi中，y在esi中。

mul:
    xorl %eax, %eax
    testl %edi, %edi  #if x == 0 return 0 //base case 1
    je done1
    testl %esi, %esi  #if y == 0 return 0 //base case 1
    je done1

    cmpl $1, %edi   #if x == 1, return y //base case 2
    je done2

    addl %esi, %eax   #add eax by y (esi) x (edi) times
    decl %edi       #decrease x by 1 each run of the function. B stays the same
    call mul

done1:
    ret

done2:              #if x == 1, return y
    movl %esi, %eax
    ret

Answer 1

它总是返回"y"，因为您将%edi与

decl %edi

直到它达到值"1“为止。然后执行以下指令序列：

call mul          # CALL -->
...
xorl %eax, %eax   # RESET accumulator value to 0
testl %edi, %edi  
je done1          # NOT TAKEN
testl %esi, %esi  
je done1          # NOT TAKEN
cmpl $1, %edi     # if x == 1, return y //base case 2
je done2          # HERE - THIS JUMP IS TAKEN
...
movl %esi, %eax   # MOVE y to return register %eax
ret               # It always returns y in %eax

要修复此情况，首先将行xorl %eax, %eax移到mul标签之前。然后删除行movl %esi, %eax以将%eax值保留为返回值。

因此，这可能是一个解决方案：

    xorl %eax, %eax   # Set result value to "0"
    # check for 0 for "y" input 
    testl %esi, %esi  # if y == 0 return 0 //base case 1
    je done1          # immediately return on zero
mul:    
    test %edi, %edi   # if x == 0, exit recursive calls
    je done1
    addl %esi, %eax   # add y (esi) to eax 
    decl %edi         # decrease x by 1 each run of the function. B stays the same
    call mul
    # PASS-THROUGH on stack unwinding
done1:
    ret

免责声明:我还没有测试过它，但是你应该知道这个想法。