栈64装配对齐 [英] Stack alignment in x64 assembly

问题描述

是怎样的价值 28小时（十进制40）是从 RSP 减去计算了以下内容：

how is the value of 28h (decimal 40) that is subtracted from rsp calculated in the following:

    option casemap:none

    includelib kernel32.lib
    includelib user32.lib

externdef MessageBoxA : near
externdef ExitProcess : near

    .data

text    db 'Hello world!', 0
caption db 'Hello x86-64', 0

    .code

main proc
    sub rsp, 28h        ; space for 4 arguments + 16byte aligned stack
    xor r9d, r9d        ; 4. argument: r9d = uType = 0
    lea r8, [caption]   ; 3. argument: r8  = caption
    lea rdx, [text]     ; 2. argument: edx = window text
    xor rcx, rcx        ; 1. argument: rcx = hWnd = NULL
    call MessageBoxA
    xor ecx, ecx        ; ecx = exit code
    call ExitProcess
main endp

    end

来自： http://www.japheth.de/JWasm/Win64_1.html

这是我的理解，我将不得不只减去 20小时，因为我使用的每一个值需要8个字节到4 20小时。那么，为什么 28小时被减去，请问是怎么导致16字节对齐？

By my understanding I would have to only subtract 20h since each value I'm using takes 8 bytes into 4 is 20h. so why is 28h being subtracted and how does that result in 16 byte alignment?

又见 64大会：是预留堆栈空间所必需的功能少于四个参数

推荐答案

我相信这是因为在此之前主被调用时，堆栈对齐。那么呼叫之后，呼叫的行为是推一个8字节的指针（呼叫者的地址）入堆栈。所以在主要开头，这是8个字​​节。16字节对齐的。因此，而不是 20小时需要 28小时，实际总数达到 28H + 8H （从呼叫）或 30小时。对准。 ：）

I believe it's because before main is called, the stack is aligned. Then after the call, the act of the call was to push an 8-byte pointer (address of the caller) onto the stack. So at the beginning of main, it's 8 bytes off of the 16-byte alignment. Therefore, instead of 20h you need 28h, bringing the actual total to 28h + 8h (from the call) or 30h. Alignment. :)

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