我创建了一个对词向量进行选择排序的过程，但有一个问题:排序完全错误。
我的向量:VET_2 DW 2, 7, 0, 1, 4, 8, 9, 3, 6, 5
; Selection Sort
SELECTION_SORT PROC
; AX = j & aux CX = i
; BX = offset/min DX = data and others
PUSH 0 ; to initialize i
MOV SI, [OFFSET VET_2]
; ----- start for(int i = 0; i < n-1; i++) -----
SLC_LOOP_FORA: ; outer loop
CALL RESET_REGIST ; reset some AX, BX, CX & DX
CALL RESET_VAR ; used to reset AUX
POP CX ; initialize i
CMP CX, 18 ; check if it's smaller than n-1 (20-2=18)
JGE SLC_FIM ; if bigger, goes to the end
MOV BX, CX ; offset receive i, the position of the smaller
; ----- start j = i+1 -----
MOV AX, CX ; AX = j.
ADD AX, 2 ; j = i+1
; ----- end j = i+1 -----
; ----- start for(int j = i+1; j < n; j++) -----
SLC_LOOP_DENTRO: ; inner loop
MOV DX, [SI+BX] ; move the smaller element to DX
MOV BX, AX ; offset receives j
CMP DX, [SI+BX] ; compare if VET_2[min]<=VET_2[j]
JL SLC_DENTRO_PULAR ; if lesser, ignore the code below
MOV BX, AX ; offset receive j, position of the smaller element
SLC_DENTRO_PULAR:
ADD AX, 2 ; inc 2 in j
CMP AX, 20 ; compare j (ax) with n
JL SLC_LOOP_DENTRO ; if it's smaller, repeat inner loop
; ----- end for(int j = n+1; j < n; j++) -----
CMP CX, BX ; compare i with the position of the smaller element
JE SLC_LOOP_FORA ; if equals, repeat outer loop, otherwise do the swap
PUSH BX ; position of the smaller element
PUSH [SI+BX] ; put vet[min] top of the stack
; ----- start aux = vet[i] -----
MOV BX, CX ; offset (BX) receives i
MOV DX, [SI+BX] ; DX receives vet_2[i]
MOV AUX, DX ; AUX receives DX
; ----- end aux = vet[i] -----
; ----- start vet[i] = vet[min] -----
POP AX ; AX receives the top of the stack (vet_2[min])
MOV [SI+BX], AX ; vet_2[i] receives DX (smaller element)
; ----- end vet[i] = vet[min] -----
; ----- start vet[min] = aux -----
POP BX ; offset (BX) receives the position of the smaller element from the stack
MOV DX, AUX ; DX receives AUX
MOV [SI+BX], DX ; vet_2[min] receives DX
; ----- end vet[min] = aux -----
ADD CX, 2 ; INC 2 on i
PUSH CX ; put in the stack
JMP SLC_LOOP_FORA repeat outer loop
; ----- end for(int i = 0; i < n-1; i++) -----
SLC_FIM: ; end the procedure
RET
SELECTION_SORT ENDP
调用选择排序过程之前:2 7 0 1 4 8 9 3 6 5
调用后选择排序过程:5 2 7 0 1 4 8 9 3 6
哪里出错了？有人可以帮助我吗？
最佳答案
问题
当不需要交换2个元素时，仍然需要提高CX中的下限。 cmp cx, bx je SLC_LOOP_FORA 忽略了这一点。此外，它还会导致堆栈不平衡(弹出的内容多于推送的内容)。
解决方案:
通过引入额外的标签可以轻松纠正问题(包括堆栈不平衡):
PUSH 0 ; to initialize i
MOV SI, OFFSET VET_2
SLC_LOOP_FORA: ; outer loop
...
CMP CX, BX ; compare i with the position of the smaller element
JE NO_SWAP ; if equals, repeat outer loop, otherwise do the swap
...
NO_SWAP:
ADD CX, 2 ; INC 2 on i
PUSH CX ; put in the stack
JMP SLC_LOOP_FORA ; repeat outer loop
考虑
; AX = j & aux CX = i ; BX = offset/min DX = data and others PUSH 0 ; to initialize i MOV SI, [OFFSET VET_2]
如果您愿意重新分配寄存器，您可以极大地简化这个程序。
使用寄存器分配，例如
; AX = j & aux DI = i
; SI = offset/min DX = data and others
PUSH 0 ; to initialize i
MOV BX, OFFSET VET_2
交换代码，例如将变成这 3 条指令:
mov ax, [bx+si]
xchg ax, [bx+di]
mov [bx+si], ax
关于assembly - 如何选择排序 - assembly 8086，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/53323996/