从程序集调用Win32的睡眠函数创建访问冲突错误 [英] Calling Win32's Sleep function from assembly creates access violation error
问题描述
我使用MASM和Visual C ++，我在x64编译。这是我的C ++代码：
// include指令
#includestdafx.h
/ / external functions
externCint Asm（）;
// main函数
int main（）
{
//调用asm
Asm（）;
// get char，return success
_getch（）;
return EXIT_SUCCESS;
}
和我的汇编代码：
extern Sleep：proc
;代码段
.code
;装配过程
Asm proc
;睡眠1秒
mov ecx，1000; ecx =睡眠时间
sub rsp，8; 8字节的影子空间
call Sleep;调用sleep
add rsp，8;摆脱阴影空间
; return
ret
Asm endp
end
使用断点定位了发生访问冲突的代码行：紧跟在我的汇编代码中的 ret 语句之后。
额外资讯：
-
> fastcall 约定将我的参数传递到
即使当我尝试调用Sleep（即使它被声明为 stdcall ），因为从我已经阅读，x64将总是使用 fastcall 约定。Sleep Sleep 相关代码。
很明显， ，我的问题是，如何摆脱访问冲突错误，我做错了什么？
编辑：
这是C ++中 Sleep（500）; 的生成程序集：
mov ecx，1F4h
call qword ptr [__imp_Sleep（13F54B308h）]
这个生成的程序集让我感到困惑...它看起来像fastcall，因为它将参数移动到ecx，但同时它不创建任何阴影空间。我不知道这是什么意思：
qword ptr [__imp_Sleep（13F54B308h）] 。
strong>再次，编辑
。 int main（）
{
000000013F991020 push rdi
000000013F991022 sub rsp，20h
000000013F991026 mov rdi，rsp
000000013F991029 mov ecx，8
000000013F99102E mov eax，0CCCCCCCCh
000000013F991033 rep stos dword ptr [rdi]
Sleep（500）; //这里是由编译器生成的asm！
000000013F991035 mov ecx，1F4h
000000013F99103A call qword ptr [__imp_Sleep（13F99B308h）]
//调用asm
Asm（）;
000000013F991040 call @ ILT + 5（Asm）（13F99100Ah）
//获取char，返回成功
_getch（）;
000000013F991045 call qword ptr [__imp__getch（13F99B540h）]
return EXIT_SUCCESS;
000000013F99104B xor eax，eax
}
解决方案如果 Asm（）是正常的C / C ++函数，例如：
void Asm（）
{
Sleep（1000）;
}
以下是我的x64编译器为它生成的：
Asm proc
push rbp;将栈重新对齐到一个16字节的边界（CALL推送8字节为调用者的返回地址）以及准备建立一个栈帧
sub rsp，32; 32字节的影子空间
mov rbp，rsp;使用当前堆栈指针
完成堆栈帧;睡眠1秒
mov ecx，1000; ecx =睡眠时间
call Sleep; call sleep
lea rsp，[rbp + 32];摆脱阴影空间
pop rbp;清除堆栈帧并将堆栈指针设置回调用方返回地址的位置
ret;返回调用者
Asm endp
调用者负责为被调用者分配参数空间，并且必须始终为4个寄存器参数分配足够的空间，即使被调用者没有这么多参数。
$ b有关x64如何使用堆栈的更多信息，请查看下面的页面：
a href =http://msdn.microsoft.com/en-us/library/ew5tede7.aspx =nofollow>堆栈分配
I'm using MASM and Visual C++, and I'm compiling in x64. This is my C++ code:
// include directive
#include "stdafx.h"
// external functions
extern "C" int Asm();
// main function
int main()
{
// call asm
Asm();
// get char, return success
_getch();
return EXIT_SUCCESS;
}
and my assembly code:
extern Sleep : proc
; code segment
.code
; assembly procedure
Asm proc
; sleep for 1 second
mov ecx, 1000 ; ecx = sleep time
sub rsp, 8 ; 8 bytes of shadow space
call Sleep ; call sleep
add rsp, 8 ; get rid of shadow space
; return
ret
Asm endp
end
Using breakpoints, I've pinpointed the line of code where the access violation occurs: right after the ret statement in my assembly code.
Extra info:
I'm using the fastcall convention to pass my parameters into Sleep (even though it is declared as stdcall), because from what I have read, x64 will always use the fastcall convention.
My Asm procedure compiles and executes with no errors when I get rid of the Sleep related code.
Even when I try to call Sleep with the stdcall convention, I still get an access violation error.
So obviously, my question is, how do I get rid of the access violation error, what am I doing wrong?
Edit:
This is the generated assembly for Sleep(500); in C++:
mov ecx,1F4h
call qword ptr [__imp_Sleep (13F54B308h)]
This generated assembly is confusing me... it looks like fastcall because it moves the parameter into ecx, but at the same time it doesn't create any shadow space. And I have no clue what this means:
qword ptr [__imp_Sleep (13F54B308h)].
And again, edit, the full disassembly for main.
int main()
{
000000013F991020 push rdi
000000013F991022 sub rsp,20h
000000013F991026 mov rdi,rsp
000000013F991029 mov ecx,8
000000013F99102E mov eax,0CCCCCCCCh
000000013F991033 rep stos dword ptr [rdi]
Sleep(500); // this here is the asm generated by the compiler!
000000013F991035 mov ecx,1F4h
000000013F99103A call qword ptr [__imp_Sleep (13F99B308h)]
// call asm
Asm();
000000013F991040 call @ILT+5(Asm) (13F99100Ah)
// get char, return success
_getch();
000000013F991045 call qword ptr [__imp__getch (13F99B540h)]
return EXIT_SUCCESS;
000000013F99104B xor eax,eax
}
解决方案 If Asm() were a normal C/C++ function, eg:
void Asm()
{
Sleep(1000);
}
The following is what my x64 compiler generates for it:
Asm proc
push rbp ; re-aligns the stack to a 16-byte boundary (CALL pushed 8 bytes for the caller's return address) as well as prepares for setting up a stack frame
sub rsp, 32 ; 32 bytes of shadow space
mov rbp, rsp ; finalizes the stack frame using the current stack pointer
; sleep for 1 second
mov ecx, 1000 ; ecx = sleep time
call Sleep ; call sleep
lea rsp, [rbp+32] ; get rid of shadow space
pop rbp ; clears the stack frame and sets the stack pointer back to the location of the caller's return address
ret ; return to caller
Asm endp
The caller is responsible for allocating space for parameters to the callee, and must always allocate sufficient space for the 4 register parameters, even if the callee doesn’t have that many parameters.
Have a look at the following page for more information about how x64 uses the stack:
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