組件程序十分簡單,組件的線程模型爲STA,代碼如下:
STDMETHODIMP CTheMath::Add(long IOp1, long IOp2, long* plResult)
{
*plResult = IOp1 + IOp2;
printf("IOp1 + IOp2 = %d", *plResult);
return S_OK;
}
調用組件的程序代碼如下:
// ATLSTLExe.cpp : 定義控制檯應用程序的入口點。
//
#include "stdafx.h"
#include <windows.h>
#include <commctrl.h>
#include <iostream>
#include "ATLSTATest.h"
#include "ATLSTATest_i.c"
using namespace std;
int ExeProc(LPVOID p)
{
CoInitializeEx(0, COINIT_MULTITHREADED);
ITheMath *ptrMath;
HRESULT hr = CoCreateInstance(CLSID_TheMath,
NULL,
CLSCTX_SERVER,
IID_ITheMath,
(void **)&ptrMath);
if FAILED(hr)
{
cout<<"Get ptrMath error!"<<endl;
CoUninitialize();
return -1;
}
cout<<"Now Begin to create file!"<<endl;
long* plresult;
ptrMath->Add(100, 200, plresult);
cout<<"100+200 = "<<*plresult<<endl;
ptrMath->Release();
CoUninitialize();
return 0;
}
int _tmain(int argc, _TCHAR* argv[])
{
int i;
i = 0;
HANDLE hr;
DWORD ThreadID;
hr = CreateThread(NULL, 0, (LPTHREAD_START_ROUTINE)ExeProc, (LPVOID)&i, 0, &ThreadID);
if (hr)
CloseHandle(hr);
MSG msg;
while (GetMessage(&msg, NULL, 0, 0))
{
TranslateMessage(&msg);
DispatchMessage(&msg);
}
return 0;
}
當上述代碼中紅色部分改爲:
CoInitializeEx(0, COINIT_APARTMENTTHREADED); //STA
程序提示如下調試信息:
加載模塊: ATLSTLExe.exe
加載模塊: ole32.dll
加載模塊: coredll.dll
Get ptrMath error!
線程 'ExeProc' (0xe7a47242) 已退出,返回值爲 -1 (0xffffffff)。
當代碼不變的時候,com爲組件創建一個STA套間,調用組件的程序中創建的線程運行在MTA套間中,對於跨套間的調用,com自動列集接口,實現同步,調試信息如下:
加載模塊: ATLSTLExe.exe
加載模塊: ole32.dll
加載模塊: coredll.dll
加載模塊: oleaut32.dll
加載模塊: atlstatest.dll
Now Begin to create file!
IOp1 + IOp2 = 300100+200 = 300
卸載模塊: atlstatest.dll
卸載模塊: oleaut32.dll
線程 'ExeProc' (0x87a1e95a) 已退出,返回值爲 0 (0x0)。
以上輸出表明程序運行正常,哪位網友能說說原因?