通过代码,我如何测试硬盘驱动器是否正在睡眠而不会唤醒它

Question

我正在构建一个小应用程序,它为我提供了磁盘上的可用空间.

我想添加一个显示磁盘状态的功能,例如,如果它处于休眠状态.操作系统是Windows.

如何才能做到这一点？当然,代码不应该唤醒磁盘找出来;)

C#中的解决方案会很好,但我猜任何解决方案都会...

谢谢你的帮助.

Answer 1

C++解决方案(调用GetDiskPowerState,它将遍历物理驱动器,直到没有更多):

class AutoHandle
{
    HANDLE  mHandle;
public:
    AutoHandle() : mHandle(NULL) { }
    AutoHandle(HANDLE h) : mHandle(h) { }

    HANDLE * operator & ()
    {
        return &mHandle;
    }

    operator HANDLE() const
    {
        return mHandle;
    }

    ~AutoHandle()
    {
        if (mHandle && mHandle != INVALID_HANDLE_VALUE)
            ::CloseHandle(mHandle);
    }
};


bool
GetDiskPowerState(LPCTSTR disk, string & txt)
{
    AutoHandle hFile = CreateFile(disk, 0, FILE_SHARE_READ|FILE_SHARE_WRITE, NULL, OPEN_EXISTING, 0, NULL);
    if (hFile && hFile != INVALID_HANDLE_VALUE)
    {
        BOOL powerState = FALSE;
        const BOOL result = GetDevicePowerState(hFile, &powerState);
        const DWORD err = GetLastError();

        if (result)
        {
            if (powerState)
            {
                txt += disk;
                txt += " : powered up\r\n";
            }
            else
            {
                txt += disk;
                txt += " : sleeping\r\n";
            }
            return true;
        }
        else
        {
            txt += "Cannot get drive ";
            txt += disk;
            txt += "status\r\n";
            return false;
        }
    }

    return false;
}

string 
GetDiskPowerState()
{
    string text;
    CString driveName;
    bool result = true;
    for (int idx= 0; result; idx++)
    {
        driveName.Format("\\\\.\\PhysicalDrive%d", idx);
        result = GetDiskPowerState(driveName, text);
    }
    return text;
}