x64程序集中的堆栈对齐

的价值是多少 28h (十进制40) rsp 计算如下:

    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

发件人: http://www.japheth.de/JWasm/Win64_1.html

根据我的理解,我只需要做减法 20h 因为我使用的每个值将8个字节分成4个 20小时 。那么为什么 28小时 被减去,这是如何导致16字节对齐的?

另请参阅 Is reserving stack space necessary for functions less than four arguments?