其實可以徹底擺脫那個static，辦法就是在直接（或者使用JITAssembler）手動生成static的代碼，跟thunk替換棧上數據一樣的道理。只是這樣生成的代碼同時與編譯器和CPU綁定了，與編譯器綁定是因為各個編譯器生成的調用成員函數的方式可能不一樣。
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#include <iostream>
#include <string>
#include <windows.h>
#include <cstdio>
using namespace std;


template<class T, class S>
inline T union_cast( S s )
{
union
{
T t;
S s;
} u;
u.s = s;
return u.t;
}

#define __CODE1( a ) \
*(proc++) = a

#define __CODE2( a1, a2 ) \
*(proc++) = a1; \
*(proc++) = a2

#define __CODE3( a1, a2, a3 ) \
*(proc++) = a1; \
*(proc++) = a2; \
*(proc++) = a3

#define __CODE4( a1, a2, a3, a4 ) \
*(proc++) = a1; \
*(proc++) = a2; \
*(proc++) = a3; \
*(proc++) = a4

#define __CODE5( a1, a2, a3, a4, a5 ) \
*(proc++) = a1; \
*(proc++) = a2; \
*(proc++) = a3; \
*(proc++) = a4; \
*(proc++) = a5

#define __CODE6( a1, a2, a3, a4, a5, a6 ) \
*(proc++) = a1; \
*(proc++) = a2; \
*(proc++) = a3; \
*(proc++) = a4; \
*(proc++) = a5; \
*(proc++) = a6

#define __CODE( n, a ) __CODE##n a

#define __PTR( p ) \
{ \
*( (void**)proc ) = union_cast<void*>(p);\
proc += sizeof(void*); \
}

#define __CALL( p ) \
{ \
*(proc++) = 0xE8; \
unsigned char* pfunc = union_cast<unsigned char*>(p);\
*( (ptrdiff_t*)proc ) = pfunc - proc - 4; \
proc += 4; \
}  回復  更多評論
  

class CTest
{
public:
CTest()
: m_value( 0x12345678 )
{
build_proc();
}

public:
LRESULT CALLBACK member_proc( HWND hwnd, UINT msg, WPARAM wparam, LPARAM lparam )
{
printf( "hwnd:%d, msg:%d, wparam:%d, lparam:%d\n", (int)hwnd, (int)msg, (int)wparam, (int)lparam );
printf( "CTest::value: 0x%x\n", m_value );
return 0;
}

void build_proc()
{
unsigned char* proc = m_proc;
///////////////////////// Prolog.
__CODE( 1, ( 0x55 ) ); // 55 push ebp
__CODE( 2, ( 0x8B, 0xEC ) ); // 8B EC mov ebp,esp
__CODE( 6, ( 0x81, 0xEC, 0xC0, 0x00, 0x00, 0x00 ) ); // 81 EC C0 00 00 00 sub esp,0C0h
__CODE( 1, ( 0x53 ) ); // 53 push ebx
__CODE( 1, ( 0x56 ) ); // 56 push esi
__CODE( 1, ( 0x57 ) ); // 57 push edi
__CODE( 6, ( 0x8D, 0xBD, 0x40, 0xFF, 0xFF, 0xFF ) ); // 8D BD 40 FF FF FF lea edi,[ebp+FFFFFF40h]
__CODE( 5, ( 0xB9, 0x30, 0x00, 0x00, 0x00 ) ); // B9 30 00 00 00 mov ecx,30h
__CODE( 5, ( 0xB8, 0xCC, 0xCC, 0xCC, 0xCC ) ); // B8 CC CC CC CC mov eax,0CCCCCCCCh
__CODE( 2, ( 0xF3, 0xAB ) ); // F3 AB rep stos dword ptr es:[edi]

////////////////////// Codes
__CODE( 3, ( 0x8B, 0x45, 0x14 ) ); // 8B 45 14 mov eax,dword ptr [ebp+14h] [lparam]
__CODE( 1, ( 0x50 ) ); // 50 push eax
__CODE( 3, ( 0x8B, 0x45, 0x10 ) ); // 8B 45 10 mov eax,dword ptr [ebp+10h] [wparam]
__CODE( 1, ( 0x50 ) ); // 50 push eax
__CODE( 3, ( 0x8B, 0x55, 0x0C ) ); // 8B 55 0C mov edx,dword ptr [ebp+0Ch] [msg]
__CODE( 1, ( 0x52 ) ); // 52 push edx
__CODE( 3, ( 0x8B, 0x45, 0x08 ) ); // 8B 45 08 mov eax,dword ptr [ebp+8] [hwnd]
__CODE( 1, ( 0x50 ) ); // 50 push eax
__CODE( 1, ( 0xB9 ) ); __PTR( this ); // B9 ?? ?? ?? ?? mov ecx, this
__CODE( 1, ( 0x51 ) ); // 51 push ecx
__CALL( &CTest::member_proc ); // E8 ?? ?? ?? ?? call CTest::member_proc

/////////////////////// Epilog.
__CODE( 1, ( 0x5F ) ); // 5F pop edi
__CODE( 1, ( 0x5E ) ); // 5E pop esi
__CODE( 1, ( 0x5B ) ); // 5B pop ebx
__CODE( 6, ( 0x81, 0xC4, 0xC0, 0x00, 0x00, 0x00 ) ); // 81 C4 C0 00 00 00 add esp,0C0h
__CODE( 2, ( 0x8B, 0xE5 ) ); // 8B E5 mov esp,ebp
__CODE( 1, ( 0x5D ) ); // 5D pop ebp
__CODE( 3, ( 0xC2, 0x10, 0x00 ) ); // C2 10 00 ret 10h

DWORD old = 0;
VirtualProtect( &m_proc, sizeof(m_proc), PAGE_EXECUTE_READWRITE, &old );
}

WNDPROC get_proc()
{
return (WNDPROC)(void*)m_proc;
}

public:
char m_proc[1024];
int m_value;
};


int main( int argc, char** argv )
{
CTest test;
WNDPROC proc = test.get_proc();
proc( (HWND)1, 2, 3, 4 );

return 0;
}  回復  更多評論
  

如果我說chromium的base庫的延遲任務更好呢？？
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@cexer
學習了  回復  更多評論
  

@WXX
有更好的很正常。  回復  更多評論