我需要获取QString(例如"A")并将此字符串转换为0x41。输入是QString,输出必须是"WORD"(微软的东西)。
在我的情况下,单词是一个无符号的短整形。我试过:
Qstring str = "A"
qDebug() << str.toLatin1().toHex().prepend("0x");
结果是很好的"0x41",但它不是一个无符号的短 int :(我试过投射,但它不起作用。
commun 答案似乎使用"toUShort",但我的代码不适用于此更改。我想用dll winuser(https://msdn.microsoft.com/en-us/library/windows/desktop/ms646270%28v=vs.85%29.aspx)模拟键盘:
此代码工作:
INPUT key;
key.type = INPUT_KEYBOARD;
key.ki.wScan = 0; // hardware scan code for key
key.ki.time = 0;
key.ki.dwExtraInfo = 0;
key.ki.wVk = 0x42; // virtual-key code for the "a" key
key.ki.dwFlags = 0; // 0 for key press
SendInput(1, &key, sizeof(INPUT));
key.ki.dwFlags = KEYEVENTF_KEYUP; // KEYEVENTF_KEYUP for key release
SendInput(1, &key, sizeof(INPUT));
没有:
QString str = "a";
bool ok;
INPUT key;
key.type = INPUT_KEYBOARD;
key.ki.wScan = 0; // hardware scan code for key
key.ki.time = 0;
key.ki.dwExtraInfo = 0;
key.ki.wVk = str.toUShort(&ok,16);
key.ki.dwFlags = 0; // 0 for key press
SendInput(1, &key, sizeof(INPUT));
key.ki.dwFlags = KEYEVENTF_KEYUP; // KEYEVENTF_KEYUP for key release
SendInput(1, &key, sizeof(INPUT));
也许只是因为短期回报 42 而不是0x42。
--溶液--
void SyncKeyboard::writeUID(QString uid)
{
qDebug() << uid;
for(int i = 0; i < uid.length(); i++)
pressKey(uid.at(i).unicode());
}
void SyncKeyboard::pressKey(ushort letter)
{
INPUT key;
key.type = INPUT_KEYBOARD;
key.ki.wScan = 0; // hardware scan code for key
key.ki.time = 0;
key.ki.dwExtraInfo = 0;
key.ki.wVk = letter;
key.ki.dwFlags = 0; // 0 for key press
SendInput(1, &key, sizeof(INPUT));
key.ki.dwFlags = KEYEVENTF_KEYUP; // KEYEVENTF_KEYUP for key release
SendInput(1, &key, sizeof(INPUT));
}
如果你只需要从A
获取一个无符号的短整数,即十六进制 ASCII 代码,你可以尝试使用 QByteArray::toUShort 代替。
ushort QByteArray::toUShort ( bool * ok = 0, int base = 16 ) const