如何编写从汇编代码调用的C函数
我需要编写一个C函数,该函数将从Linux内核中的汇编代码调用。如何编写从汇编代码调用的C函数
应该考虑哪些特殊问题?
我有一些想法,但有谁能够提供更多的细节:
(1)调用约定
确保在装配呼叫者和在C握手言和被叫方好。但是我应该使用哪种调用约定?我如何声明c函数并在汇编代码中声明它?
(2)保护的寄存器
某些寄存器应的调用之前被保存在组件。我大致记得他们是eax,ecx和edx。但我不必保存它们,除非我需要在调用C函数之前引用旧值,对吧?
还有什么其他问题?
由于你的任务标记为linux-kernel
请看下面的内核头文件:arch/x86/include/asm/calling.h。
3 x86 function call convention, 64-bit:
4 -------------------------------------
5 arguments | callee-saved | extra caller-saved | return
6 [callee-clobbered] | | [callee-clobbered] |
7 ---------------------------------------------------------------------------
8 rdi rsi rdx rcx r8-9 | rbx rbp [*] r12-15 | r10-11 | rax, rdx [**]
9
10 (rsp is obviously invariant across normal function calls. (gcc can 'merge'
11 functions when it sees tail-call optimization possibilities) rflags is
12 clobbered. Leftover arguments are passed over the stack frame.)
13
14 [*] In the frame-pointers case rbp is fixed to the stack frame.
15
16 [**] for struct return values wider than 64 bits the return convention is a
17 bit more complex: up to 128 bits width we return small structures
18 straight in rax, rdx. For structures larger than that (3 words or
19 larger) the caller puts a pointer to an on-stack return struct
20 [allocated in the caller's stack frame] into the first argument - i.e.
21 into rdi. All other arguments shift up by one in this case.
22 Fortunately this case is rare in the kernel.
23
24 For 32-bit we have the following conventions - kernel is built with
25 -mregparm=3 and -freg-struct-return:
26
27 x86 function calling convention, 32-bit:
28 ----------------------------------------
29 arguments | callee-saved | extra caller-saved | return
30 [callee-clobbered] | | [callee-clobbered] |
31 -------------------------------------------------------------------------
32 eax edx ecx | ebx edi esi ebp [*] | <none> | eax, edx [**]
33
34 (here too esp is obviously invariant across normal function calls. eflags
35 is clobbered. Leftover arguments are passed over the stack frame.)
36
37 [*] In the frame-pointers case ebp is fixed to the stack frame.
38
39 [**] We build with -freg-struct-return, which on 32-bit means similar
40 semantics as on 64-bit: edx can be used for a second return value
41 (i.e. covering integer and structure sizes up to 64 bits) - after that
42 it gets more complex and more expensive: 3-word or larger struct returns
43 get done in the caller's frame and the pointer to the return struct goes
44 into regparm0, i.e. eax - the other arguments shift up and the
45 function's register parameters degenerate to regparm=2 in essence.
太棒了!这让我很困惑。你能告诉我你是如何找到这么好的文件的吗?一个愚蠢的问题。但严重的是,我查了几个小时,但没有找到这样的文件。顺便说一句:你能建议一些可以指导内核编程的书籍/链接,尤其是低级时尚吗?非常感谢! – Infinite 2012-04-08 15:33:34
嗯,我只知道在哪里看:)说到这些书你可以在这里搜索类似的问题。除此之外,你总是可以看看Linux内核源码本身。另外,欢迎提出问题。 – 2012-04-08 16:15:58
x86_64调用约定的主要参考是http://www.x86-64.org/documentation/abi.pdf(“ABI” - _Application Binary Interface_)。它列出了调用约定和组合数据('struct')以及'原始'类型布局/大小/对齐约束。 Linux内核在64位x86中使用的约定与此非常接近。然而,在32位版本中,它们与用户空间ABI(如果您好奇的话,在网上搜索_gabi.pdf_)完全不同,因为使用了'gcc -mregparm' - 这就是内核源代码派上用场的地方。 – 2012-05-31 17:13:27
“我需要编写一个C函数,它将从Linux内核的汇编代码中调用。” - 什么东西阻止你? – 2012-04-08 08:00:24
@MitchWheat具体来说,在声明c函数时我不知道“asmlinkage”的确切用法。我必须使用它吗?还有其他的选择吗?含义是什么? – Infinite 2012-04-08 08:02:52
首先 - 尽量不要这样做。 Linux在几个地方使用汇编,而且它们都很棘手。我认为这些问题取决于它是什么汇编代码。在中断处理程序的早期阶段或早期启动阶段(使用C的两个地方)运行是两个不同的事情。 – ugoren 2012-04-08 09:34:16