Why is RegOpenKeyEx() returning error code 2 on Vista 64bit?

Question

I was making the following call:

result = RegOpenKeyEx(key, s, 0, KEY_READ, &key);

(C++, Visual Studio 5, Vista 64bit).

It is failing with error code 2 ("File not found") even though "regedit" shows that the key exists. This code has always worked on 32bit XP. Why is it "file not found" when it clearly is there?

Solution 1

I discovered that I could solve my problem using the flag: KEY_WOW64_64KEY , as in:

result = RegOpenKeyEx(key, s, 0, KEY_READ|KEY_WOW64_64KEY, &key);

For a full explanation: 32-bit and 64-bit Application Data in the Registry

Solution 2

On a Windows 64-bit system the Registry is actually divided into two parts. One section is used by 64-bit processes, and one part by 32-bit processes.

For example, if a 32-bit application programatically writes to what it believes is HKLM\SOFTWARE\Company\Application, it's actually redirected by the WoW64-layer to HKLM\SOFTWARE\Wow6432Node\Company\Application.

So when you run your 32-bit application and call RegOpenKeyEx it's actually working against the Wow6432Node\ folder, and not the regular \SOFTWARE node.

Solution 3

You have to compile with "Use Multi-Byte Character Set" or cast string in code to (LPWSTR)