我想在两个进程之间创建一个共享内存。因此,我简单地复制从Microsoft页面的剪断:无法映射共享内存
片段1:
#define BUF_SIZE 256
TCHAR szName[] = TEXT("Global\\MyFileMappingObject");
TCHAR szMsg[] = TEXT("Message from first process.");
void initSharedMem() {
HANDLE hMapFile;
LPCTSTR pBuf;
hMapFile = CreateFileMapping(
INVALID_HANDLE_VALUE, // use paging file
NULL, // default security
PAGE_READWRITE, // read/write access
0, // maximum object size (high-order DWORD)
BUF_SIZE, // maximum object size (low-order DWORD)
szName); // name of mapping object
if (hMapFile == NULL) {
MessageBox(0, "Could not create file mapping object", "Error", 0);
return;
}
pBuf = (LPTSTR)MapViewOfFile(hMapFile, // handle to map object
FILE_MAP_ALL_ACCESS, // read/write permission
0,
0,
BUF_SIZE);
if (pBuf == NULL) {
MessageBox(0, "Could not map view of file", "Error", 0);
CloseHandle(hMapFile);
return;
}
CopyMemory((PVOID)pBuf, szMsg, (_tcslen(szMsg) * sizeof(TCHAR)));
_getch();
UnmapViewOfFile(pBuf);
CloseHandle(hMapFile);
MessageBox(0, "Done init shared mem", "Done", 0);
return;
}
片段2(其他工序):
#define BUF_SIZE 256
TCHAR szName[] = TEXT("Global\\MyFileMappingObject");
TCHAR szMsg[] = TEXT("Message from first process.");
void readSharedMem() {
HANDLE hMapFile;
LPCTSTR pBuf;
hMapFile = OpenFileMapping(
FILE_MAP_ALL_ACCESS, // read/write access
FALSE, // do not inherit the name
szName); // name of mapping object
if (hMapFile == NULL) {
MessageBox(0, L"Error", L"Could not open file mapping object", 0);
return;
}
pBuf = (LPTSTR)MapViewOfFile(hMapFile, // handle to map object
FILE_MAP_ALL_ACCESS, // read/write permission
0,
0,
BUF_SIZE);
if (pBuf == NULL) {
MessageBox(0, L"Error", L"Could not map file", 0);
CloseHandle(hMapFile);
return;
}
MessageBox(NULL, pBuf, TEXT("Process2"), MB_OK);
UnmapViewOfFile(pBuf);
CloseHandle(hMapFile);
return;
MessageBox(0, L"Done", L"SharedMemoryDone", 0);
}
我调用过程A中的第一功能,并获得完成的消息。但是当我之后调用readSharedMem函数时,我收到错误消息“无法打开文件映射对象”。
我在这里做错了什么?
当您尝试打开共享内存部分时,您的进程仍然活着吗?此外,当'OpenFileMapping'失败时,你会得到什么样的错误代码? – IInspectable
提示:永远不要复制/粘贴代码。理想情况下,您应该重新输入代码,查找您不了解的任何内容。这就是你学习新概念的方法。 – CodeMouse92
是的,两者都还活着,错误代码是2. – QDA