无法将用户输入的数字与nasm elf64进行比较 [英] can't compare user input with number, nasm elf64

问题描述

我发誓今天我已经读了20多个页面，从NASM的手册到大学的Wikipedia指南，以及介于两者之间的所有内容，但我对此无能为力，我编写了一个程序来比较用户输入用0或1表示，然后根据该值进行操作(一旦在Assembly中挂起它们，我可能应该使用数组)，但这现在就可以了.

I swear I've read more than 20 pages today, from NASM's manual to Universities' guides to Wikipedia to everything in between but I just can't wrap my head around this, I wrote a single program to compare the user input with either a 0 or a 1 and then act based on that (I should probably use an array once I get the hang of them in Assembly), but this will do for now.

问题是，我的支票永远无效，它们总是直接指向错误标签，我看着

Problem is, my checks never work, they always go straight to the err label, I looked at x86 NASM Assembly - Problems with Input which seemed similar but it wasn't quite the same, I really don't need to store the user's input, just check what it was and react on it.

这是我代码的简化版本，应该在输入两个连续的0后退出，显然我无法对其进行测试，因为我无法弄清楚用户输入了什么.

Here's a reduce version of my code, it's supposed to exit after two consecutive 0's have been entered, clearly I haven't been able to test it, since I can't figure out what the user entered.

很抱歉，这是一个愚蠢的问题，但是本周大会对我来说是最好的.

I'm sorry if this is a silly question but this week Assembly has gotten the best of me.

; constants
section .data
  lblZero:      db    'Zero';
  lblOne:       db    'One ';
  lblNumLength: db    0x4;
  tmp:          db    0;

; code
section .text
  global _start

; linker needs this, 'main'
_start:

loop:
  ; user was already prompted for a single digit
  ; store user's input ; read
  mov rax, 0 ;
  mov rbx, 19 ;
  mov rcx, tmp ;
  mov rdx, 10 ;
  syscall
  ; series of IFs
  cmp   rcx, 0    ; is input 0? 00 exits the program
  je    isZero
  cmp   rcx, 1    ; is input 1?
  je    isOne  
  jmp   exit  

; user typed 0
isZero:
  inc   rcx               ; flag for 0
  cmp   rcx, 2            ; checking if this is the 2nd zero
  je    exit              ; if so, we are outta here
  mov   rsi, lblZero ;
  mov   rcx, -1 ;
  jmp  print ;

; user typed 1
isOne:
  mov   rsi, lblOne ;
  mov   rcx, -1 ;
  jmp  print ;

; prints the string into the screen ; sys_write
print:
  mov   rax, 1 ;           
  mov   rdi, 1 ;         
  mov   rdx, lblNumLength ;       
  syscall
  jmp   loop

; displays an error message
err:
  ; sys_write, not relevant to the Q     
  syscall
  jmp loop

; shutsdown program ; sys_write ; sys_exit
exit:
  ; not relevant to the Q, code always ends here

我在这里阅读 http://www.cin .ufpe.br/〜if817/arquivos/asmtut/index.html#stack ，输入不是一个真正的整数，而是一个字符，因此我尝试创建仅存储"1"或"0"但似乎无济于事.

I read here http://www.cin.ufpe.br/~if817/arquivos/asmtut/index.html#stack that the input is not truly an int, but a character, so I tried creating variables that just store '1' or '0' but nothing seems to do the trick.

我是Assembly的超级新手，所以如果我在这里做的很愚蠢，请指出.

I'm a super new to Assembly so if I'm doing really stupid here, I would appreciate if you point it out.

推荐答案

假设您尝试在64位AMD64 linux上运行此程序，则需要确保使用正确的方法将参数传递给内核(调用惯例).

Assuming you're trying to run this on 64 bit AMD64 linux, you need to make sure you use the right way of passing parameters to the kernel (the calling convention).

对于Linux/x64上的64位应用程序，如下所示:

For a 64 bit app on linux/x64, it is as follows:

• rax -系统调用ID
• rdi -参数1
• rsi -参数2
• rdx -参数3
• r10 -参数4
• arg 5和6在 r9 和 r8
中
• rcx 和 r11 的值不会在整个系统调用中保留，但所有其他寄存器都不会被破坏.

• rax - system call id
• rdi - argument 1
• rsi - argument 2
• rdx - argument 3
• r10 - argument 4
• arg 5 and 6 are in r9 and r8
• rcx and r11 values are not preserved across syscall, but all other registers will not be clobbered.

此外，Paul R的答案也是正确的. ASCII表示的'0'不是十进制的零.另一个陷阱是，标准输入默认情况下在缓冲模式下运行，因此，除非您按Enter键，否则实际上不会获得任何数据.下面的代码将两个字节读入tmp缓冲区(由rsi指向)，并将第一个字节与ascii-零进行比较.第二个字节是换行符，我们对此并不特别感兴趣.

In addition, Paul R's answer is also correct. The ASCII representation of '0' is not decimal zero. Another gotcha is that the stdin operates in buffered mode by default, so you won't actually get any data until you press enter. The code below reads two bytes into the tmp buffer (pointed to by rsi) and compares the first with ascii-zero. The second byte is a newline character that we're not particularly interested in.

; constants
section .data
  lblZero:      db    'Not zero, try again', 0xa;
  tmp:          db    0,0;
; code

section .text
BITS 64
  global _start
_start: 
  mov rax, 0   ; sys_read(int fd, void *ptr, int count)
  mov rdi, 0   ; 0 = stdin 
  mov rsi, tmp ; @ tmp
  mov rdx, 2   ; 2 bytes (one for our number, another to store newline)
  syscall
  cmp byte[rsi], '0' ; is input '0' (or decimal 48) ? 
  je  done
  mov rax, 1            ; sys_write(int fd, void *ptr, int count)
  mov rdi, 1            ; 1 = stdout
  mov rsi, lblZero      ; @lblZero
  mov rdx, 20           ; 20 bytes
  syscall
  jmp _start

done:   
  mov rax, 60 ; sys_exit
  mov rdi, 0
  syscall

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