blocks|key|3153206|text|您的引用来自文档的“调用约定”部分。至少，如果您不从汇编代码中调用其他函数，则不必担心这一点。如果你这样做了，那么你必须尊重“红色区域”和堆栈对齐考虑因素，这是你引用的建议旨在确保的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3153207|编辑：this+post澄清了“红色区域”和“阴影空间”之间的区别。|offset|length|3153208|entityMap|0|LINK|mutability|MUTABLE|url|http://eli.thegreenplace.net/2011/09/06/stack-frame-layout-on-x86-64/^0|0|3|9|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@$D|P|E|Q|1|R]]|A|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|M]]]]

Your quote is from the "calling convention" part of the documentation. At the very least, you do not have to worry about this if you do not call other functions from your assembly code. If you do, then you must respect, among other things, "red zone" and stack alignment considerations, that the recommendation you quote is intended to ensure.

EDIT: <a href="http://eli.thegreenplace.net/2011/09/06/stack-frame-layout-on-x86-64/" rel="noreferrer">this post</a> clarifies the difference between "red zone" and "shadow space".

blocks|key|2938843|text|在更多地使用它并阅读文档之后，需要为您调用的任何函数保留这32个字节。如果您的函数与示例一样简单，并且不调用其他函数，则不必保留此空间。但是，您调用的任何函数都可能使用这32个字节，因此如果您不保留它们，函数可能会|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2938844|另外，你的函数可能依赖于调用你的函数的堆栈上有32个字节可用，如果它遵循ABI的话。通常，此32字节区域用于保存将在函数中更改的寄存器，以便您可以在返回之前恢复它们的值。我认为这是出于性能目的，选择32字节足以使大多数叶函数(不调用其他函数)不需要保留任何堆栈空间，并且在堆栈上有临时空间来保存寄存器并在返回之前恢复它们。举个例子：|2938845|调用函数：|2938846|CallingFunction:
++push+rbp
++mov+rbp,+rsp
++sub+rsp,+40++//+$20+bytes+we+want+to+use+at+[rbp%2B30],
+++++++++++++++//+plus+$20+bytes+for+calling+other+functions
+++++++++++++++//+according+to+windows+ABI+spec
++mov+rcx,+[rsi%2B10]+++++//+parameter+1+(xmm0+if+non-int)
++mov+rdx,+10+++++++++++//+parameter+2+(xmm1+if+non-int)
++movss+xmm2,+[rsi%2B28]++//+parameter+3+(r8+if+int)
++mov+r9,+[rsi%2B64]++++++//+parameter+4+(xmm3+if+non-int)
++call+MyFunction
++//+...+do+other+stuff
++add+rsp,+40+++++++++++//+free+space+we+reserved
++pop+rbp
++xor+rax,rax
++ret|code-block|syntax|javascript|2938847|调用的函数|2938848|CalledFunction:
++push+rbp++++++//+standard
++mov+rbp,+rsp++//+standard

++//+should+do+'sub+rsp,+20'+here+if+calling+any+functions
++//+to+give+them+a+free+scratch+area

++//+[rbp]+is+saved+rbp
++//+[rbp%2B8]+is+return+address
++//+[rbp%2B10]+to+[rbp%2B2f]+are+the+0x20+bytes+we+can
++//+++++safely+modify+in+this+function,+this+could
++//+++++be+pushed+higher+if+the+function+had+more+than+4
++//+++++parameters+and+some+had+to+be+passed+on+the+stack
++//+++++or+if+returning+a+structure+or+something+that+needs
++//+++++more+space.++In+these+cases+the+CALLER+would+have
++//+++++allocated+more+space+for+us

++//+the+main+reason+for+the+0x20+is+so+that+we+can+save+
++//+registers+we+want+to+modify+without+having+to+allocate
++//+stack+space+ourselves
++mov+[rbp%2B10],+rsi+//+save+rsi+in+space+allocated+by+caller
++mov+[rbp%2B18],+rdi+//+save+rdi+in+space+allocated+by+caller
++mov+rsi,+[rcx%2B20]
++mov+rdi,+[rsi%2B48]
++add+rdi,+[rsi%2B28]
++mov+rax,+rdi
++mov+rdi,+[rbp%2B18]+//+restore+changed+register
++mov+rsi,+[rbp%2B10]+//+restore+changed+register
++pop+rbp
++ret|2938849|原始答案|offset|length|style|BOLD|2938850|我只是在不知情的情况下遇到了这种情况，似乎是这样的。例如，GetAsyncKeyState中的前两条指令覆盖了返回地址之上的堆栈，该堆栈位于应该为被调用者保留的0x20字节区域中，以供被调用者用于参数：|2938851|user32.GetAsyncKeyState++-+mov+[rsp%2B08],rbx
user32.GetAsyncKeyState%2B5-+mov+[rsp%2B10],rsi
user32.GetAsyncKeyState%2BA-+push+rdi
user32.GetAsyncKeyState%2BB-+sub+rsp,20|2938852|entityMap^0|0|0|0|0|0|0|0|4|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|12|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|13|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|15|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|16|8|@$Q|17|R|18|S|T]]|9|@]|A|$]]|$1|U|3|V|5|6|7|19|8|@]|9|@]|A|$]]|$1|W|3|X|5|H|7|1A|8|@]|9|@]|A|$I|J]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

After playing with this more and reading the docs, the 32 bytes need to be reserved for any function that you call. If your function is as simple as the example and you don't call other functions, you don't have to reserve this space. Any function you call however may use this 32 bytes so if you do not reserve them the function may
Also your function may rely on there being 32 bytes available on the stack from the function that called yours if it's following the ABI. Commonly this 32 byte area is used to save registers that will be changed in your function so you can restore their values before returning. I think is is for performance purposes, 32 bytes being chosen as enough to make it so most
leaf functions (functions that don't call others) don't need to reserve any stack space, and have temporary room on the stack to save registers and restore them before returning. Take this example:
Calling Function:
<pre><code>CallingFunction:
 push rbp
 mov rbp, rsp
 sub rsp, 40 // $20 bytes we want to use at [rbp+30],
 // plus $20 bytes for calling other functions
 // according to windows ABI spec
 mov rcx, [rsi+10] // parameter 1 (xmm0 if non-int)
 mov rdx, 10 // parameter 2 (xmm1 if non-int)
 movss xmm2, [rsi+28] // parameter 3 (r8 if int)
 mov r9, [rsi+64] // parameter 4 (xmm3 if non-int)
 call MyFunction
 // ... do other stuff
 add rsp, 40 // free space we reserved
 pop rbp
 xor rax,rax
 ret
</code></pre>
Called Function
<pre><code>CalledFunction:
 push rbp // standard
 mov rbp, rsp // standard

 // should do 'sub rsp, 20' here if calling any functions
 // to give them a free scratch area

 // [rbp] is saved rbp
 // [rbp+8] is return address
 // [rbp+10] to [rbp+2f] are the 0x20 bytes we can
 // safely modify in this function, this could
 // be pushed higher if the function had more than 4
 // parameters and some had to be passed on the stack
 // or if returning a structure or something that needs
 // more space. In these cases the CALLER would have
 // allocated more space for us

 // the main reason for the 0x20 is so that we can save 
 // registers we want to modify without having to allocate
 // stack space ourselves
 mov [rbp+10], rsi // save rsi in space allocated by caller
 mov [rbp+18], rdi // save rdi in space allocated by caller
 mov rsi, [rcx+20]
 mov rdi, [rsi+48]
 add rdi, [rsi+28]
 mov rax, rdi
 mov rdi, [rbp+18] // restore changed register
 mov rsi, [rbp+10] // restore changed register
 pop rbp
 ret
</code></pre>
Original answer
I just ran into this not knowing and it seems to be the case. The first two instructions in GetAsyncKeyState for instance overwrite the stack above the return address in the 0x20 byte area you're supposed to reserve for the callee to use for parameters:
<pre><code>user32.GetAsyncKeyState - mov [rsp+08],rbx
user32.GetAsyncKeyState+5- mov [rsp+10],rsi
user32.GetAsyncKeyState+A- push rdi
user32.GetAsyncKeyState+B- sub rsp,20
</code></pre>

Just started learning x64 assembly and I have a question about functions, arguments, and the stack. As far as I understand it, the first four arguments in a function get passed to rcx, rdx, r8, and r9 registers (and xmm0-xmm3 for floats) in Windows. So a trivial addition function with four parameters would looks like this:

<pre><code>add:
 mov r10, rcx
 add r10, rdx
 add r10, r8
 add r10, r9
 mov rax, r10
 ret
</code></pre>

However, I've come across <a href="http://msdn.microsoft.com/en-us/magazine/cc300794.aspx" rel="nofollow noreferrer">documentation that mentions this</a>:

<blockquote>
 At a minimum, each function must reserve 32 bytes (four 64-bit values) on the stack. This space allows registers passed into the function to be easily copied to a well-known stack location. The callee function isn't required to spill the input register params to the stack, but the stack space reservation ensures that it can if needed.
</blockquote>

So, do I have to reserve stack space even if the functions I'm making take four parameters or less, or is it just a recommendation?

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

Windows

刚开始学习x64汇编，我有一个关于函数、参数和堆栈的问题。据我所知，函数中的前四个参数被传递给Windows中的rcx、rdx、r8和r9寄存器(对于浮点数则是xmm0-xmm3 )。因此，一个带有四个参数的简单加法函数将如下所示：add:   mov r10, rcx   add r10, rdx   add r10...

问是否需要为少于四个参数的函数预留堆栈空间？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问是否需要为少于四个参数的函数预留堆栈空间？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问是否需要为少于四个参数的函数预留堆栈空间？
EN