我被要求在 NASM Ubuntu 中创建一个冒泡排序程序。这是代码：
section .data
i db 0 ; Value to be incremented
question db 'Enter a number: ' ; Prompt
questionLen equ $-question
newLine db 10, 10, 0 ; New blank line
newLineLen equ $-newLine
section .bss
num resb 5 ; Array of size 5
counter resb 1 ; Value to be incremented
counter2 resb 1 ; Value to be incremented
temp resb 1
temp2 resb 1
section .text
global _start
_start:
mov esi, 0
getInput:
mov eax, 4
mov ebx, 1
mov ecx, question ; Prints the question
mov edx, questionLen
int 80h
add byte[i], 30h ; I'll retain this expression, since the program experienced an error
; when this expression is deleted
sub byte[i], 30h ; Converts the increment value to integer
mov eax, 3
mov ebx, 0
lea ecx, [num + esi] ; Element of the array
mov edx, 2
int 80h
inc esi
inc byte[i]
cmp byte[i], 5 ; As long as the array hasn't reached the size of 5,
jl getInput ; the program continues to ask input from the user
mov esi, 0
mov byte[i], 0
mov edi, 0 ; Index of the array
bubble_sort:
mov byte[counter], 0
mov byte[counter2], 0
begin_for_1:
mov al, 0
mov al, [counter] ; Acts as the outer for loop
cmp al, 5
jg printArray ; Prints the sorted list when the array size has reached 5
begin_for_2:
mov edi, [counter2] ; Acts as the inner for loop
cmp edi, 4
jg end_for_2
mov bl, 0 ; Acts as the if statement
mov cl, 0
mov bl, [num + edi]
mov cl, [num + edi + 1]
mov byte[temp], cl ; This is the same as if(a[j] > a[j + 1]){...}
cmp bl, [temp]
jg bubbleSortSwap
return:
inc edi ; Same as j++
jmp begin_for_2 ; Goes out of the inner for loop
end_for_2:
inc byte[counter] ; Same as i++
jmp begin_for_1 ; Goes out of the outer for loop
bubbleSortSwap:
mov [num + edi + 1], bl
mov [num + edi], cl ; The set of statements is the same as swap(&a[j], &a[j + 1]);
jmp return
printArray:
mov eax, 4
mov ebx, 1
mov ecx, [num + esi] ; Prints one element at a time
mov edx, 1
int 80h
inc esi
inc byte[i]
cmp byte[i], 5
jl printArray ; As long as the array size hasn't reached 5, printing continues
mov eax, 4
mov ebx, 1
mov ecx, newLine ; Displays a new blank line after the array
mov edx, newLineLen
int 80h
mov eax, 1 ; Exits the program
mov ebx, 0
int 80h
但唯一的问题是，它不能打印其余的迭代，因为它只打印第一次迭代，如下所示：
Enter a number: 7
Enter a number: 1
Enter a number: 4
Enter a number: 3
Enter a number: 5
17435
我要输出的是数组输入和最终输出，从第一次迭代到最后一次。