c++ - MapViewOfFile 在 32 位和 64 位进程之间共享

Question

我正在尝试在 64 位进程中对已映射到另一个 32 位进程内存的文件使用 MapViewOfFile。它失败并给我一个“拒绝访问”错误。这是一个已知的 Windows 限制还是我做错了什么？相同的代码适用于 2 个 32 位进程。

代码大概是这样的:

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); 

Answer 1

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:我刚刚整理了一个有效的演示(至少对我而言):

#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;
}

32 位或 64 位可执行文件的任何组合似乎都可以正常工作。

Edit2:但是请注意，这是纯粹的演示级代码。例如，每个共享对象的名称通常应包含一个 GUID 字符串，以确保不会与其他程序发生意外冲突。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细节。