为什么 RegOpenKeyEx() 在 Vista 64 位上返回错误代码 2?

2021-12-18 00:00:00 winapi windows-vista c++ registry wow64

我拨打了以下电话:

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

(C++、Visual Studio 5、Vista 64 位).

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

即使regedit"显示密钥存在，它也失败并显示错误代码 2(找不到文件").此代码始终适用于 32 位 XP.明明有文件，为什么会找不到文件"?

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?

推荐答案

我发现我可以使用标志来解决我的问题:KEY_WOW64_64KEY，如下所示:

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);

完整说明:32 位和 64-注册表中的位应用程序数据

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