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 f

Answer 1

yes，win7 64B,add further flag KEY_WOW64_64KEY ,it will work. if not work, refer to http://msdn.microsoft.com/en-us/library/ms724897(v=VS.85).aspx