我有两个函数从输入中读取整数 x 和 y。
产品返回 x * y
power 返回 x ^ y，但是它使用递归和乘积来计算它。所以 x 是“基数”，y 是“指数”。
他们从 C++ 调用：
int a, b, x, y;
a = product(x, y);
b = power(x, y);
这是汇编。我让产品工作，但是在电源方面遇到问题，因为我不确定从中调用产品的语法/方法/约定（并调用自身进行递归）。编辑：必须使用递归。
global product
global power
section .text
product:
push ebp
mov ebp, esp
sub esp, 4
push edi
push esi
xor eax, eax
mov edi, [ebp+8]
mov esi, [ebp+12]
mov [ebp-4], edi
product_loop:
add [ebp-4], edi
mov eax, [ebp-4]
sub esi, 1
cmp esi, 1
jne product_loop
product_done:
pop esi
pop edi
mov esp, ebp
pop ebp
ret
power:
push ebp
mov ebp, esp
sub esp, 4
push edi
push esi
push ebx
xor eax, eax
mov edi, [ebp+8]
mov esi, [ebp+12]
;;;
check:
cmp esi, 1 ; if exp < 1
jl power_stop
recursion: ; else (PLEASE HELP!!!!!!!!)
; eax = call product (base, (power(base, exp-1))
power_stop:
mov eax, 1 ; return 1
power_done:
push ebx
pop esi
pop edi
mov esp, ebp
pop ebp
ret
编辑：我的解决方案！
power:
; Standard prologue
push ebp ; Save the old base pointer
mov ebp, esp ; Set new value of the base pointer
sub esp, 4 ; make room for 1 local variable result
push ebx ; this is exp-1
xor eax, eax ; Place zero in EAX. We will keep a running sum
mov eax, [ebp+12] ; exp
mov ebx, [ebp+8] ; base
cmp eax, 1 ; n >= 1
jge L1 ; if not, go do a recursive call
mov eax, 1 ; otherwise return 1
jmp L2
L1:
dec eax ; exp-1
push eax ; push argument 2: exp-1
push ebx ; push argument 1: base
call power ; do the call, result goes in eax: power(base, exp-1)
add esp, 8 ; get rid of arguments
push eax ; push argument 2: power(base, exponent-1)
push ebx ; push argument 1: base
call product ; product(base, power(base, exponent-1))
L2:
; Standard epilogue
pop ebx ; restore register
mov esp, ebp ; deallocate local variables
pop ebp ; Restore the callers base pointer.
ret ; Return to the caller.