Question

我正在尝试在Linux上编写一个32位x86 NASM程序，它从文本文件中获取一个字符串，编码将字符串中的所有字符向上移动1并将printf移出到控制台。几乎一切都有效，除了用于进行移位和.rot132的循环printf。每当调用.rot132时，它都会给出segmentation_fault（核心转储）。我想当我做mov ecx, [eax]时会引起某种错误。我怎样才能克服这一部分？

;;sddddddddddddddddgsd

[SECTION .data] ; Section containing initialized data

WriteCode db "w",0
OpenCode db "r",0
Filename db "text.txt",0
fileFmt: dd "%c\n",10,0
fileFmt1: dd "%s",10,0

[SECTION .bss] ; Section containing uninitialized data
TextLenght EQU 72 ; Define length of a line of  text data
Text resb TextLenght ; Reserve space for disk-based help text line
BUFSIZE EQU 64 ; Define length of text line buffer 
Buff resb BUFSIZE+5 ; Reserve space for a line of text

[SECTION .text] ; Section containing code
;; These externals are all from the glibc standard C library:
extern fopen
extern fclose
extern fgets
extern fprintf
extern printf
extern sscanf
extern time

global main ; Required so linker can find entry point

main:

diskhelp:
        mov  ebx, Filename ; push file name to ebx
        push OpenCode ; Push pointer to open-for-read code “r“
        push ebx ; Pointer to name of help file is passed in ebx
        call fopen ; Attempt to open the file for reading
        add esp,8 ; Clean up the stack
        cmp eax,0 ; fopen returns null if attempted open failed
        jne .disk ; Read help info from disk file...
        ret

.disk:  mov ebx,eax ; Save handle of opened file in ebx

.rdln:  push ebx ; Push file handle on the stack
        push dword TextLenght ; Limit line length of text read
        push Text ; Push address of text line buffer
        call fgets ; Read a line of text from the file
        add esp,12 ; Clean up the stack
        ;cmp eax,0 ; A returned null indicates error or EOF
        ;jle .rot13 ; If we get 0 in eax, close up & return

        ;push Text ; Push address of help line on the stack
        ;call printf ; Call printf to display help line
        ;add esp,4 ; Clean up the stack

.rot131: ; initial shift and test, this work
        mov edx, 0    ; our counter
        mov eax, Text ; move string into eax
        mov ecx, [eax]; move first char in string into ecx
        add ecx, 1    ; shift the char up by 1 
        push ecx      ; push to print
        push fileFmt
        call printf
        add esp, 8   ; clear the stack
        inc edx      ; increase the counter

.rot132: 
        inc eax     ; shift address of eax into next char
        mov ecx, [eax] ; move the char into ecx, replace old char; error ??
        add ecx, 1     ; shift the char by 1
        push ecx        ; print
        push fileFmt
        call printf
        add esp, 8      ; clear the stack
        inc edx         ;incrase counter
        cmp edx,4       ; stop loop after edx = 4 
        jne .rot132


        push ebx ; Push the handle of the file to be closed
        call fclose ; Closes the file whose handle is on the stack
        add esp,4 ; Clean up the stack

        ret ; Go home

gdb debug：反汇编.rot132：

   0x08048559 <+0>:     inc    %eax
   0x0804855a <+1>:     mov    (%eax),%ecx
   0x0804855c <+3>:     add    $0x1,%ecx
   0x0804855f <+6>:     push   %ecx
   0x08048560 <+7>:     push   $0x804a035
   0x08048565 <+12>:    call   0x80483b0 <printf@plt>
   0x0804856a <+17>:    add    $0x8,%esp
   0x0804856d <+20>:    inc    %edx
   0x0804856e <+21>:    cmp    $0x4,%edx
   0x08048571 <+24>:    jne    0x8048559 <rot132>
   0x08048573 <+26>:    push   %ebx
   0x08048574 <+27>:    call   0x80483d0 <fclose@plt>
   0x08048579 <+32>:    add    $0x4,%esp
   0x0804857c <+35>:    ret
   0x0804857d <+36>:    xchg   %ax,%ax
   0x0804857f <+38>:    nop

Answer 1

fileFmt需要db，而不是dd。您正在定义双字符。

请注意，此代码不如编译器生成的代码。只关注组件语言，因为关键循环是配置文件顶部明显的CPU瓶颈，而且你已经在使用一个好的算法。

一个例子是矩阵乘法。它是embarassingly parallel，因此强制编译器使用可用于执行操作的最有效方法通常是有益的，例如使用向量扩展（如SSE）。

编译器非常好，你通常会编写＃34;汇编语言＆＃34;使用内在函数，例如SSE或AVX内在函数，它允许编译器担心数据流和寄存器分配，并允许您专注于指令本身。

另请注意Michael Petch的好评：允许来电隐私eax，ecx，edx。您可以相信他们会保留ebx，esi，edi，ebp，esp。这意味着您需要停止对变量使用ecx和edx，并使用保留的内容，例如ebx，edi，esi，{{1 }}

汇编语言修改输入字符串错误

1 个答案: