OS X-x64:堆栈未对齐16字节的错误 [英] OS X - x64: stack not 16 byte aligned error

问题描述

我知道OS X是16字节堆栈对齐，但是我真的不明白为什么它在这里引起错误.

I know that OS X is 16 byte stack align, but I don't really understand why it is causing an error here.

我在这里要做的就是将对象大小(为24)传递给％rdi，然后调用malloc.这个错误是否意味着我必须要32个字节?

错误消息是:

libdyld.dylib`stack_not_16_byte_aligned_error: -> 0x7fffc12da2fa< +0>:movdqa％xmm0，(％rsp) 0x7fffc12da2ff< +5>:int3

libdyld.dylib`stack_not_16_byte_aligned_error: -> 0x7fffc12da2fa <+0>: movdqa %xmm0, (%rsp) 0x7fffc12da2ff <+5>: int3

libdyld.dylib`_dyld_func_lookup: 0x7fffc12da300< +0>:pushq％rbp 0x7fffc12da301< +1>:movq％rsp，％rbp

libdyld.dylib`_dyld_func_lookup: 0x7fffc12da300 <+0>: pushq %rbp 0x7fffc12da301 <+1>: movq %rsp, %rbp

这是代码:

Object_copy:
    pushq %rbp
    movq %rbp, %rsp

    subq $8, %rsp
    movq %rdi, 8(%rsp)          # save self address
    movq obj_size(%rdi), %rax   # get object size
    imul $8, %rax          
    movq %rax, %rdi 
    callq _malloc             <------------------- error in this call

    # rsi old object address
    # rax new object address
    # rdi object size, mutiple of 8

    # rcx temp reg

    # copy object tag
    movq 0(%rsi), %rcx
    movq %rcx, 0(%rax)

    # set rdx to counter, starting from 8
    movq $8, %rdx

    # add 8 to object size, since we are starting from 8
    addq $8, %rdi

    start_loop:
        cmpq %rdx, %rdi
        jle end_loop

        movq (%rdx, %rsi, 1), %rcx
        movq %rcx, (%rdx, %rax, 1)

        addq $8, %rdx
        jmp start_loop

    end_loop:
        leave 
        ret



Main_protoObj:
    .quad    5                          ; object tag
    .quad    3                          ; object size
    .quad    Main_dispatch_table        ; dispatch table

_main:
    leaq Main_protoObj(%rip), %rdi
    callq Object_copy                # copy main proto object
    subq $8, %rsp                    # save the main object on the stack
    movq %rax, 8(%rsp)
    movq %rax, %rdi                 # set rdi point to SELF
    callq Main_init
    callq Main_main

    addq $8, %rsp                    # restore stack

    leaq _term_msg(%rip), %rax
    callq _print_string

推荐答案

就像您说的那样，MacOS X具有16字节的堆栈对齐方式，这意味着计算机希望堆栈上的每个变量都以一个整数倍的字节开始.当前堆栈指针中的16个.

Like you said, MacOS X has a 16 byte stack alignment, which means that the machine expects each variable on the stack to start on a byte that is a multiple of 16 from the current stack pointer.

当堆栈未对齐时，这意味着我们开始尝试从16个字节的窗口的中间读取变量，并且通常会遇到分段错误.

When the stack is misaligned, it means we start trying to read variables from the middle of that 16 byte window and usually end up with a segmentation fault.

在代码中调用例程之前，需要确保堆栈正确对齐；在这种情况下，意味着基址指针寄存器可以被16整除.

Before you call a routine in your code, you need to make sure that your stack is aligned correctly; in this case, meaning that the base pointer register is divisible by 16.

subq $8, %rsp               # stack is misaligned by 8 bytes
movq %rdi, 8(%rsp)          #
movq obj_size(%rdi), %rax   #
imul $8, %rax               #
movq %rax, %rdi             #
callq _malloc               # stack is still misaligned when this is called

要解决此问题，您可以将subq %rsp的大小改为16(而不是8).

To fix this, you can subq the %rsp by something like 16 instead of 8.

subq $16, %rsp               # stack is still aligned
movq %rdi, 16(%rsp)          #
...                          #
callq _malloc                # stack is still aligned when this is called, good

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