A（应该是）简单的程序与NASM doesnt't工作 [英] A (should be) simple program with nasm doesnt't work

问题描述

我是新来这个论坛。
我有高级语言（还真有点）一点经验。近一个月前我认为这将是一个好主意，看看如何装配在Linux上选择NASM（IA-32），我开始从一个教程的学习后，曾如此。
现在，结束后，我试着写一个简单的程序，你在哪里得到的电脑打印的100号（1 2 4 8 16 ......）列表，但我甚至不能得到它的权利。
我得到这样的输出：
1PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP（继续）...

I'm new to this forum. I have a little experience with high-level languages (really little). Nearly one month ago I thought it would be a good idea to see how assembly worked so after choosing nasm (IA-32) on linux I started learning from a tutorial. Now, after ending it, I tried to write a simple program where you get the computer to print a list of 100 number (1 2 4 8 16...) but I couldn't even get it right. I get this output: 1PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP(continues)...

该计划是这样的：

section .text
    global main
main:
    mov word [num], '1'
    mov ecx, 100
doubl:
    push ecx ; to push the loop counter

    mov eax, 4
    mov ebx, 1
    mov ecx, num
    mov edx, 1
    int 0x80

    sub ecx, 30h
    add ecx, ecx   ; shl ecx, 1
    add ecx, 30h
    mov [num], ecx   ; deleting this line I get  11111111111111111...

    pop ecx  ; to pop the loop counter
    loop doubl
exit:
    mov eax, 1
    int 0x80    
section .bss
num resw 2

它看起来像错误是在双打的数量或存储在变量民的一个组成部分，但我不明白为什么会发生，以及如何解决它。

It looks like the error is in the part that doubles the number or the one that stores it in the variable 'num', yet I don't understand why it happens and how to solve it.

有人可以解释我时准确使用方括号？是否有一个规则还是什么？本教程将其称为有效地址它看起来像我不得不使用支架时，我想移动（或做的东西）变量的内容，而不是做给变量的地址。然而，我很困惑的。我看到它适用于：

By the way can someone explain me when to use the square brackets exactly? Is there a rule or something? The tutorial calls it "effective address" and it looks like I have to use the brackets when I want to move (or do something with) the content of a variable instead of doing it to the variable's address. Yet I'm quite confused about it. I see it used in:

mov ebx, count
inc word [ebx]
mov esi, value
dec byte [esi]

不过，是不是很明显，一个人想增加寄存器的内容（因为它没有一个地址（或不是吗？）??

But isn't it obvious that one wants to increment the content of the register (since it doesn't have an address (or does it?) ??

我希望有人能帮助我。预先感谢您的回答。

I hope someone can help me. Thanks in advance for your answers.

推荐答案

您数字将快速增长不仅仅是一个单一的数字更大。什么你应该做的是在 NUM 一个整数，而不是一个字符，然后将其转换整数为字符串，可以用 SYS_WRITE打印。

Your numbers will quickly grow larger than just a single digit. What you ought to be doing is have an integer in num rather than a character, and then convert that integer into a string that you can print with sys_write.

下面是执行转换的一种方式：

Here's one way of doing the conversion:

; Input:
; EAX = integer value to convert
; ESI = pointer to buffer to store the string in (must have room for at least 10 bytes)
; Output:
; EAX = pointer to the first character of the generated string
int_to_string:
  add esi,9
  mov byte [esi],0    ; String terminator

  mov ebx,10         
.next_digit:
  xor edx,edx         ; Clear edx prior to dividing edx:eax by ebx
  div ebx             ; eax /= 10
  add dl,'0'          ; Convert the remainder to ASCII 
  dec esi             ; store characters in reverse order
  mov [esi],dl
  test eax,eax            
  jnz .next_digit     ; Repeat until eax==0
  mov eax,esi
  ret

您可以使用这样的：

Which you can use like this:

mov dword [num],1
...
mov eax,[num]
mov esi,buffer
call int_to_string
; eax now holds the address that you pass to sys_write
...
num resd 1
buffer resb 10

您数量加倍可以简化为 SHL DWORD [NUM]，1

Your number-doubling can be simplified to shl dword [num],1

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