NASM 你能解释一下为什么我的代码会返回一个分段错误吗？

Question

我的代码 returns 第二次调用子例程 prntlf 时出现分段错误。

_start:
mov     ecx, msg
call prntlf

call prntlf ; Here is where the issue is

经过实验，我发现只有当我没有将ecx的值设置回我想要它打印的字符串时才会出现，但我想知道的是为什么我必须这样做那。鉴于我从保存的堆栈中弹出寄存器的值，ecx 不应该仍然保留要打印的有效字符串吗？

完整来源：

文件hello.asm:

; RUN WITH `nasm -f elf32 hello.asm -o hello.o && ld -m elf_i386 hello.o -o hello && ./hello`

%include 'subroutines.inc'

section .data
    msg         db      "Hello, World!", 0x0
    msg2         db      "Goodbye, Moon!", 0x0
    linefeed    db      0xA, 0xD

section .text
    global _start:

_start:
    mov     ecx, msg
    call prntlf

    call prntlf

    jmp     end

文件：subroutines.inc

    ; subroutines.inc


;===============================================================================
    ; getstrlen | Gets String Length and pushes the value to register edx
    ;===============================================================================
    getstrlen:
        push    eax
        push    ebx
        mov     eax, ebx

    findnterm:
        cmp     byte [eax], 0
        jz      gotstrlen
        inc     eax
        jmp     findnterm

    gotstrlen:
        sub     eax, ebx
        mov     edx, eax
        pop     eax
        pop     ebx
        ret

    ;===============================================================================
    ; printstr | Prints a String using a dynamic algorithm to find null terminator
    ;===============================================================================
    printstr:
        push    eax
        push    ebx
        push    edx

        mov     ebx, ecx
        call getstrlen

        mov     ebx, 0x01
        mov     eax, 0x04
        int 0x80

        pop     eax
        pop     ebx
        pop     edx
        ret

    ;===============================================================================
    ; prntlf | Prints a String and appends a Linefeed.
    ;===============================================================================
    prntlf:
        push    eax
        push    ebx
        push    ecx
        push    edx

        mov     eax, ecx

    movetoendloop:
        cmp     byte [eax], 0
        jz      donemoving
        inc     eax
        jmp     movetoendloop

    donemoving:
        call printstr
        mov     ecx, linefeed
        call printstr

        pop     eax
        pop     ebx
        pop     ecx
        pop     edx
        ret

    ;===============================================================================
    ; end | calls kernel and tells it to End the program
    ;===============================================================================
    end:
        mov     eax, 0x01
        mov     ebx, 0x00
        int 0x80

Answer 1

栈是后进先出的结构，意思是你最后入栈的东西会最先出栈，不管你入栈的是哪个寄存器。

prntlf:
    push    eax
    push    ebx
    push    ecx
    push    edx

    ...

    pop     edx       ; Pop them back from the stack in the reverse order in which you pushed them.
    pop     ecx
    pop     ebx
    pop     eax
    ret

NASM 你能解释一下为什么我的代码会返回一个分段错误吗？

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