【汇编语言实战】实现输出集合{1，2，...,n}全排列

#include <iostream>
using namespace std;
int A[101];
void print_permutation(int n, int *A, int cur) {
    if (cur == n) { //递归边界
        for (int i = 0; i < n; i++) {
            printf("%d ", A[i]);
        }
        printf("\n");
    } else {
        for (int i = 1; i <= n; i++) { //尝试在A[cur]中填各种整数i
            bool ok = true; //检查i是否被用过
            for (int j = 0; j < cur; j++) {
                if (A[j] == i) {
                    ok = false; //如果i已经在A[0]~A[cur-1]出现过，则不能再选
                }
            }
            if (ok) {
                A[cur] = i;
                print_permutation(n, A, cur + 1); //递归调用
            }
        }
    }
}
int main() {
    int n;
    scanf("%d", &n);
    print_permutation(n, A, 0);
    return 0;
}

INCLUDE Irvine32.inc
.data
    arr dd 120 dup(?)
    ; arr dd 1,2,3,4,5,6,7,8,9,10
.code
main PROC
    ; call readInt
    ; mov edx,eax
    mov eax,3 ;eax代表输入的数
    push eax ;eax:count
    mov edx,offset arr ;edx:arr
    push edx
    call Assign
    ; mov edx,offset arr 
    ; push edx ;edx:arr
    ; ; mov eax,edx
    ; push eax ;eax:count
    ; call output
    mov ecx,0
    push eax
    push ecx
    push edx
    call FullPermutation
    exit
main ENDP
FullPermutation PROC
    push ebp
    mov ebp,esp
    pushad
    mov edx,[ebp+8] ;edx:arr
    mov esi,[ebp+12] ;esi:begin
    mov edi,[ebp+16] ;edi:end
    mov ecx,0
    cmp esi,edi
    jl else1
again:
    cmp ecx,edi
    jge final
    mov eax,[edx+ecx*4]
    call writedec
    inc ecx
    jmp again
else1:
    mov ecx,esi
again1:
    cmp ecx,edi ;ecx:i
    jge final
    cmp esi,ecx
    je next1
    push dword ptr [edx+esi*4]
    push dword ptr [edx+ecx*4]
    pop dword ptr [edx+esi*4]
    pop dword ptr [edx+ecx*4]
next1:
    push edi
    mov eax,esi
    inc eax
    push eax
    push edx
    call FullPermutation
   
    cmp esi,ecx
    je next2
    push dword ptr [edx+esi*4]
    push dword ptr [edx+ecx*4]
    pop dword ptr [edx+esi*4]
    pop dword ptr [edx+ecx*4]
next2:
    inc ecx
    jmp again1
final:
    call crlf
    popad
    pop ebp
    ret 12
FullPermutation ENDP
Assign PROC
    push ebp
    mov ebp,esp
    pushad
    mov ecx,[ebp+12] ;ecx:count
    mov edx,[ebp+8] ;edx:arr
    mov edi,0
again:
    cmp edi,ecx
    jge final
    mov eax,edi
    inc eax
    mov [edx+edi*4],eax
    inc edi
    jmp again
final:
    popad
    pop ebp
    ret 8
Assign ENDP
output PROC
    push ebp
    mov ebp,esp
    pushad
    mov edx,[ebp+12]; edx:arr
    mov ecx,[ebp+8]; ecx:count
    mov ebx,0
next:
    cmp ebx,ecx
    jge final
    mov eax,[edx+ebx*4]
    ; mov eax,ebx
    call writeint
    inc ebx
    jmp next
final:
    popad
    ret 8
output ENDP
END main