MAX_PATH 是否足以保存来自 GetSystemDirectory() 的路径？
Is MAX_PATH enough to hold the path from GetSystemDirectory()?
根据我的理解，该路径将是一个单字母（驱动程序），后跟 "\WINDOWS\SYSTEM32"，因此 MAX_PATH 足以容纳由 [=14 填充的路径=]。所以这样做是安全的：
TCHAR dir[MAX_PATH] = {0};
if(GetSystemDirectory(dir, sizeof(dir) / sizeof(*dir)) == 0) {
// check for GetLastError()
}
还是我遗漏了什么？
为 GetSystemDirectory (which is ShGetFolderPath) 的推荐替代方案提供的文档对其 pszPath 参数说明如下：
A pointer to a null-terminated string of length MAX_PATH which will
receive the path. If an error occurs or S_FALSE is returned, this
string will be empty. The returned path does not include a trailing
backslash. For example, "C:\Users" is returned rather than
"C:\Users\".
所以，是的，MAX_PATH 将是一个足够大的缓冲区大小。
根据我的理解，该路径将是一个单字母（驱动程序），后跟 "\WINDOWS\SYSTEM32"，因此 MAX_PATH 足以容纳由 [=14 填充的路径=]。所以这样做是安全的：
TCHAR dir[MAX_PATH] = {0};
if(GetSystemDirectory(dir, sizeof(dir) / sizeof(*dir)) == 0) {
// check for GetLastError()
}
还是我遗漏了什么？
为 GetSystemDirectory (which is ShGetFolderPath) 的推荐替代方案提供的文档对其 pszPath 参数说明如下：
A pointer to a null-terminated string of length MAX_PATH which will receive the path. If an error occurs or S_FALSE is returned, this string will be empty. The returned path does not include a trailing backslash. For example, "C:\Users" is returned rather than "C:\Users\".
所以，是的，MAX_PATH 将是一个足够大的缓冲区大小。