从 32 位 c++ 程序读取 64 位注册表？

Question

I'm trying to read the registry from my c++ program, this works fine when I run it in 64-bit mode, but when I switch over to 32bit mode it does not. Now I've learned that there are 2 separate parts of the registry, the 64 and 32-bit parts. I read that I should use KEY_WOW64_64KEY but I can't figure out how. The following code is what i had:

    char value[255];
    DWORD BufferSize = BUFFER;
    RegGetValue(HKEY_LOCAL_MACHINE, "SOFTWARE\\Microsoft\\Cryptography", "MachineGuid", RRF_RT_ANY, NULL, (PVOID)&value, &BufferSize);
    std::wcout << value << std::endl;
    system("pause");

And it works fine, but when I switch over to 32bit, it doesn't. So I tried the following: replacing RRF_RT_ANY with the RRF_RT_ANY | KEY_WOW64_32KEY

    char value[255];
    DWORD BufferSize = BUFFER;
    RegGetValue(HKEY_LOCAL_MACHINE, "SOFTWARE\\Microsoft\\Cryptography", "MachineGuid", RRF_RT_ANY | KEY_WOW64_32KEY, NULL, (PVOID)&value, &BufferSize);
    std::wcout << value << std::endl;
    system("pause");

But this also doesn't work, what is the correct way to apply this?

It returns a value of "ÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌÌ"

Answer 1

According to the documentation :

RRF_SUBKEY_WOW6464KEY

If lpSubKey is not NULL, open the subkey that lpSubKey specifies with the KEY_WOW64_64KEY access rights.

So you can modify the code as:

RegGetValue(HKEY_LOCAL_MACHINE, L"SOFTWARE\\Microsoft\\Cryptography",L"MachineGuid", RRF_RT_ANY | RRF_SUBKEY_WOW6464KEY, NULL, (PVOID)&value, &BufferSize);

Then it works for me.