x64 调用约定(堆栈)和可变参数 [英] x64 calling convention (stack) and varargs

问题描述

我已经阅读了 Microsoft 的文档，但计划是如此尴尬，我想我会仔细检查以确保我理解正确...

I've read Microsoft's documentation, but the scheme is so awkward, I thought I'd double-check to make sure I'm understanding it correctly...

我的理解是传递参数的通用方法是这样的:

My understanding is the generic method by which parameters are passed is this:

--- bottom of stack ---
(return address)
[shadow space for arg 1]
[shadow space for arg 2]
[shadow space for arg 3]
[shadow space for arg 4]
arg N
arg N - 1
arg N - 2
...
arg 6
arg 5
---- top of stack -----

在实现 va_arg 时看起来很尴尬，这样......这真的正确吗?

It seems so awkward when implementing va_arg and such... is this actually correct?

推荐答案

正确的图是

--- Bottom of stack ---    RSP + size     (higher addresses)
arg N
arg N - 1
arg N - 2
...
arg 6
arg 5
[shadow space for arg 4]
[shadow space for arg 3]
[shadow space for arg 2]
[shadow space for arg 1]
(return address)
---- Top of stack -----    RSP            (lower addresses)
[grows downward]

返回地址在栈顶(最近被压栈)，后面是前四个参数的影子空间，后面是参数 5 及以后.

The return address is at the top of the stack (most recently pushed), followed by shadow space for the first four parameters, followed by parameters 5 and onward.

参数从右到左入栈:最后一个参数(N)先入栈，所以最靠近栈底.

The parameters are pushed right to left: The last parameter (N) is pushed first, so it is closest to the bottom of the stack.

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