32位和64位进程之间共享的MapViewOfFile

Question

32位和64位进程之间共享的MapViewOfFile

c++windows64-bit32bit-64bitmemory-mapped-files

7

我正在尝试在一个64位进程中使用MapViewOfFile函数，该函数用于将一个已经映射到另一个32位进程内存的文件映射到当前进程。但是这时会失败并显示“访问被拒绝”的错误信息。这是Windows的已知限制还是我的操作有误？相同的代码在两个32位进程中正常工作。

代码大致如下：

hMapFile = OpenFileMapping(FILE_MAP_ALL_ACCESS, FALSE, szShmName);
if (NULL == hMapFile)
{   /* failed to open - create new (this happens in the 32 bit app) */
   SECURITY_ATTRIBUTES  sa;
   sa.nLength = sizeof(SECURITY_ATTRIBUTES);
   sa.bInheritHandle = FALSE;  
   /* give access to members of administrators group */
   BOOL success = ConvertStringSecurityDescriptorToSecurityDescriptor(
            "D:(A;OICI;GA;;;BA)",
            SDDL_REVISION_1,
            &(sa.lpSecurityDescriptor),
            NULL);
   HANDLE hShmFile = CreateFile(FILE_XXX_SHM, 
            FILE_ALL_ACCESS, 0, 
            &sa, 
            OPEN_ALWAYS, 0, NULL);

   hMapFile = CreateFileMapping(hShmFile, &sa, PAGE_READWRITE, 
            0, 
            SHM_SIZE, 
            szShmName);

   CloseHandle(hShmFile);
}

// this one fails in 64 bit app
pShm = MapViewOfFile(hMapFile, FILE_MAP_ALL_ACCESS, 0, 0, SHM_SIZE);

- MK.

你检查过路径是否被重定向到VirtualStore目录了吗？Process Monitor可能会有所帮助。 - bk1e

1个回答

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- Jerry Coffin · Accepted Answer

当您在32位应用程序中调用CreateFile时，您将传递0作为共享参数，这意味着不允许共享。将其更改为FILE_SHARE_READ | FILE_SHARE_WRITE可能是朝着正确方向迈出的一步。

编辑：我刚刚组合了一个演示，它可以正常工作（至少对我来说）。

#include <windows.h>
#include <iostream>

static const char map_name[] = "FileMapping1";
static const char event1_name[] = "EventName1";
static const char event2_name[] = "EventName2";

int main() { 
    HANDLE mapping = OpenFileMapping(FILE_MAP_ALL_ACCESS, FALSE, map_name);

    if (NULL == mapping) {
        std::cout << "Calling CreateFile\n";
        HANDLE file = CreateFile("MappedFile", 
            FILE_ALL_ACCESS, 
            FILE_SHARE_READ | FILE_SHARE_WRITE,
            NULL, 
            OPEN_ALWAYS, 
            0, 
            NULL);
        std::cout << "Creating File mapping\n";
        mapping = CreateFileMapping(file, NULL, PAGE_READWRITE, 0, 65536, map_name);

        std::cout << "Closing file handle\n";
        CloseHandle(file);
    }

    std::cout << "Mapping view of file\n";
    char *memory = (char *)MapViewOfFile(mapping, FILE_MAP_ALL_ACCESS, 0, 0, 65536);
    if (memory == NULL) {
        std::cerr << "Mapping Failed.\n";
        return 1;
    }
    std::cout << "Mapping succeeded\n";

    HANDLE event = CreateEvent(NULL, false, false, event1_name);

    if (GetLastError()==ERROR_ALREADY_EXISTS) {
        std::cout <<"Waiting to receive string:\n";
        WaitForSingleObject(event, INFINITE);
        std::cout << "Received: " << memory;
        HANDLE event2 = CreateEvent(NULL, false, false, event2_name);
        SetEvent(event2);
    }
    else {
        char string[] = "This is the shared string";
        std::cout << "Sending string: " << string << "\n";
        strncpy(memory, string, sizeof(string));
        SetEvent(event);
        HANDLE event2 = CreateEvent(NULL, false, false, event2_name);
        WaitForSingleObject(event2, INFINITE);
    }   
    return 0;
}

任何32位或64位可执行文件的组合似乎都可以正常工作。

编辑2：请注意，这仅是演示级别的代码。例如，每个共享对象的名称通常应包含GUID字符串，以确保不会与其他程序意外冲突。我还跳过了相当多的错误检查，更不用说这段代码没有完成任何有用的事情了。