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

c++ winapi windows-vista registry wow64
Tim Cooper picture Tim Cooper · Oct 31, 2008 · Viewed 55.3k times · Source

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?

Answer

Tim Cooper picture Tim Cooper · Oct 31, 2008

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

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