无法将数字转换为字符串汇编器 NASM

无法在 Assembler NASM 上将数字转换为带有 ASCII 字符的字符串。即 123 -> "1","2","3"。这个数字是无符号的。走这条路

1. 首先，我向子例程传递了一个参数——需要处理的数字。
2. 然后他每次除以1000000，如果private里面有东西，那么我们把这个bit写入栈中。之前，我把 0 放在堆栈上，并与它进行比较，以防有人输入 00456 而不是 456。如果堆栈中的第一个数字不等于 0，我们将其写入堆栈。
3. 我们将除数减少 10。也就是说，每次我们减少排放量，其余部分将在排放量中包含或不存在。例如，205 - 200/100 = 2 余数 5。如果除数（数字校验）为 1，那么我们只需将余数写入堆栈并继续从堆栈中读取。
4. 我把所有的东西都出栈了，同时我设置了栈顶不是ebp，也就是每次递增ecx的时候是不是到了末尾，这样我以后就知道字符串的长度了。这种方法使我不必跟踪我在堆栈中放入了多少数字。
5. 复制同时改变记录方向的线。由于栈中的123写成123，所以我会从栈中得到321，也就是我先开始读，改加48，从最后写。

我已经审查了所有内容一千次，似乎我已经考虑了所有内容，但是代码不起作用。我不会问这个问题，因为我只是在学习并且不知道如何使用调试器。非常感谢所有回复的人。我不想退出，问题没有解决，就像经常发生的那样。

这是代码本身：

global start_create

section .bss
string resb 7                           ;variable parameter
string2 resb 7

section .data
string2_end equ string2
string2_length dd 0
division10 dd 10                        ;we will add the counter to get a end of string: how many elements we will have after loop


section .text
start_create:   push ebp                ;save the ebp 
                mov ebp, esp            ;get the link to esp. ebp stores just lik of its value
                mov dword [ebp-4], 0    ;set initial value to check if stack has any value
                xor edx, edx            ;edx:eax/ecx ; set 0 in edx because edx:eax will be the divisible
                mov eax, [ebp+8]        ;get the number(parameter) transfered by the calling procedure
                mov ecx, 1000000        ;set the digit to divide - divisor - YOU LATER CAN SET ANY NUMBER

division:       div ecx                 ;get the quotient of the number
                test eax, eax           ;check if the quotient zero
                jnz put_not_zero        ;if it is not so - push the value in stack - otherwise check the first value in stack 
                cmp dword [ebp-4], 0    ;check if initial value on stack is 0, it means we will not put the number in stack
                jnz push_in_stack       ;if first argument is not 0, push 0 in stack 0!+0 12...+0  
                jmp next_digit          ;if the first argument is 0 

put_not_zero:   cmp dword [ebp-4], 0    ;
                jnz push_in_stack
                mov [ebp-4], eax        ;clear the first initial value with normal value

push_in_stack:  push eax                ;save the digit in stack

next_digit:     cmp ecx, 1             ; check if divisor is 10 to finish the loop
                jne continue_division 
                push edx                ;if divisor is 10 we will not continue loop. We will put the remainder in stack 
                jmp read_to_create_string

continue_division:  
                mov ebx, edx            ;move the remainder to ebx to save it
                mov eax, ecx            ;decrement divisor by 10 lower
                xor edx, edx            ;zero the value of the edx, because edx is part of division operation 
                div dword [division10]  
                mov ecx, eax            ;set the new digit divisor
                mov eax, ebx            ;get the remainder for division operation 
                jmp division  

read_to_create_string:
                mov ecx, 0 
read_from_stack: 
                cmp esp, ebp            ;check if we go to a start of the adress of procedure [return adress - 4] = ebp, or we do not have string
                je create_ascii_symbols
                pop edx
                mov [string + ecx], dl  ;write the in string 
                inc ecx                 ;increment ecx to write next number from stack to memory
                jmp read_from_stack

create_ascii_symbols:
                mov [string2_length], ecx ; change value to strore length of taken string to display it later
                add [string2_end], cl    ;string2_end == string2 we add cl instead because we add byte to byte, and in ecx we have not so big number              
                mov esi, string
                mov edi, string2_end
add48_to_ASCII: cld                     ;we do not set value in ecx - counter because we have it afer previously
                lodsb                   ;load number to create ascii symbol
                add eax, 48             ;
                std                     ;change direction to write not 3,2,1 to 1,2,3 . To write from the end
                stosb                   ;
                loop add48_to_ASCII     ;repeat the action several times pointed in ecx - length of the string

                                        ;in string we have string of numbers in ASCII 
display:        mov eax, 4
                mov ebx, 1
                mov ecx, string
                mov edx, string2_length
                int 80h


                mov esp, ebp 
                pop ebp
                ret