assembly 为什么不打印？NASM x86编码

这是它应该打印的内容
输入以某种语言编写的字符串：“如果土拨鼠会扔木头，它会扔多少木头？”
a：2 B：0 c：10 d：6 e：0 f：1 g：0 h：6 i：1 j：0 k：4 l：2 m：1 n：0 o：11 p：0 q：0 r：0 s：0 t：0 u：7 v：0 w：6 x：0 y：0 z：0观察员
然而我的只是打印：输入以某种语言编写的字符串：“如果土拨鼠会扔木头，它会扔多少木头？”
a：2 B：0 c：10 d：6 e：0 f：1 g：0 h：6 i：1 j：0 k：4 l：2 m：1 n：0 o：11 p：0 q：0 r：0 s：0 t：0 u：7 v：0 w：6 x：0 y：0 z：0

%include "asm_io.inc"

section .data
    msg     db      "Enter a string written in some language: ", 0

section .bss
    counters resd    26 ; an array of 26 counters, each storing a 4-byte value
    sorted   resd    26 ; an array of 26 letters, sorted by decreasing number of occurrences
    input    resd   1   ; to store the input character

section .text
    global asm_main

asm_main:
    ; setup
    enter   0, 0
    pusha

    ; print prompt
    mov eax, msg
    call print_string

    ; loop to count characters
    mov ebx, counters
loop:
    call read_char ; read input character
    cmp  eax, 10 ; check if end of input (newline character)
    je end ; if newline character, exit loop
    cmp eax, 97 ; check if lowercase
    jl upper_case ; if less than 97, ignore
    cmp eax, 122 ; check if lowercase
    jg not_alpha ; if greater than 122, check uppercase
    sub eax, 97 ; ASCII value subtract by 97 to get index
    jmp update_counter

lowercase_counter:
    cmp eax, 0
    je next
    add ebx, 4
    dec eax
    jmp lowercase_counter

upper_case:
    cmp eax, 65 ; check if uppercase
    jl not_alpha ;if less than 65, ignore it
    cmp eax, 90 ; check if uppercase
    jg not_alpha ; if greater than 90, ignore it
    sub eax, 65 ; ASCII value subtract by 65 to get index
    jmp update_counter

uppercase_counter:
    cmp eax, 0
    je next
    add ebx, 4
    dec eax
    jmp uppercase_counter

not_alpha:
    ; ignore non-letter characters
    jmp loop

update_counter:
    inc dword [ebx+eax*4]
    mov ebx, counters
    jmp loop

next:
    mov ebx, counters
    jmp loop

end:
    ; print character counts
    mov ebx, counters
    mov ecx, 97
loop2:
    cmp ecx, 122
    jg end2
    mov eax, ecx
    call print_char
    mov eax, 58
    call print_char ; print the colon character
    mov eax, [ebx]
    call print_int
    mov eax, 32
    call print_char
    inc ecx
    add ebx, 4
    jmp loop2

 ; sort letters by decreasing count
    mov eax, sorted
    mov ebx, counters
    mov ecx, 26

outer_loop:
    mov edx, [ebx]
    mov [eax], edx
    add eax, 4
    add ebx, 4
    loop outer_loop

    mov eax, sorted
    mov ebx, 24

sort_loop:
    mov ecx, 0

inner_loop:
    mov edx, [eax]      ; move the value of the first element of the array into edx
    mov ebx, [eax + 4]  ; move the value ofthe second element of the array into ebx
    cmp edx, ebx ; compare the two values
    jge skip_swap ; if the first value is greater than or equal to the second value, skip the swap
    mov [eax], ebx ; move the second value to the first element of the array
    mov [eax + 4], edx ; move the first value to the second element of the array

skip_swap:
    add eax, 4 ; move to the next pair of elements in the array
    loop inner_loop ; repeat until all pairs have been compared and swapped (if necessary)
    dec ebx             ; move to the next element in the array, excluding the last element since it is already sorted
    cmp ebx, 0          ; check if we have reached the beginning of the array
    jge sort_loop       ; if not, repeat the sorting process for the remaining unsorted elements

   ; print sorted letters and their counts
    call print_nl
    mov eax, sorted
    mov ebx, counters
    mov ecx, 26

print_sorted:
    cmp ecx, 0
    jle end2
    mov edx, [eax]
    add eax, 4
    sub ecx, 1
    cmp edx, 0
    je print_sorted   ; skip printing if the count is 0
    add edx, 97
    mov eax, edx
    call print_char
    mov ebx, 58      ; store the ASCII code for ":" in ebx
    call print_char  ; print ":"
    mov eax, [ebx + (edx - 97) * 4] ; retrieve the count using the letter's ASCII value
    call print_int
    call print_nl
    jmp print_sorted

    

end2:
    call print_nl
    popa ; clean up
    mov eax, 0 ; clean up
    leave ; clean up
    ret ; clean up

为什么不打印？
您的程序没有输出“ocuwhdklamifzyxvtsrqpnjgeb”，因为您在打印完计数器之后已经跳到了末尾。

; print character counts
    mov ebx, counters
    mov ecx, 97
loop2:
    cmp ecx, 122
    jg end2

现在，上面的错误是一种祝福，因为如果排序代码被允许执行，它将被证明包含一个无限循环。您初始化ECX=0，然后使用LOOP指令。这段代码将循环40亿次！
此外，您打算将EBX寄存器用于两个完全不同的目的：一次作为某种计数器，一次作为数组元素。这正是崩溃的组成部分。为什么不使用更多不同的寄存器呢？你还有ESI，EDI和EBP可以玩。
您试图使用计数器对数组副本进行降序冒泡排序。如果不采取预防措施，您将失去与字母表中字母的任何连接！
就我个人而言，我会将字符包含在要排序的记录中的某个地方，但只使用实际计数来确定顺序：

; sort letters by decreasing count
  mov  edi, sorted
  mov  esi, counters
  mov  ecx, 00006100h     ; CH='a'
copy_loop:
  mov  eax, [esi]         ; Counter [0,255]
  add  esi, 4
  or   eax, ecx           ; Combine character in AH with count in AL 
  mov  [edi], eax
  add  edi, 4
  add  ecx, 00000100h
  cmp  ecx, 00007A00h     ; CH='z'
  jbe  copy_loop

  mov  ecx, 25            ; ECX is outer loop counter
outer_loop:
  mov  esi, sorted
  mov  edx, ecx           ; EDX is inner loop counter
inner_loop:
  mov  eax, [esi]
  mov  ebx, [esi + 4]
  cmp  al, bl             ; Compare the two COUNTS ONLY
  jge  skip_swap
  mov  [esi], ebx         ; But swap the WHOLE RECORD
  mov  [esi + 4], eax
skip_swap:
  add  esi, 4
  dec  edx
  jnz  inner_loop
  dec  ecx
  jnz  outer_loop

现在，sorted 处的内存如下所示：

11, 'o', 0, 0, 10, 'c', 0, 0, 7, 'u', 0, 0, ...

因此，您需要编写 * 一个简单的循环 *，从每个双字中检索第二个字节，然后将其打印出来。
不要使用当前版本的 print_sorted，它包含许多错误！

assembly 为什么不打印？NASM x86编码

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