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(shù)據(jù)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(diào)用成員函數(shù)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(fù)  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(fù)  更多評論
  

如果我說chromium的base庫的延遲任務(wù)更好呢？？
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@cexer
學(xué)習(xí)了  回復(fù)  更多評論
  

@WXX
有更好的很正常。  回復(fù)  更多評論