【问题标题】:Mapping of virtual keycode to unicode influenced by writing to std::cout?虚拟键码到 unicode 的映射受写入 std::cout 的影响?
【发布时间】:2014-09-17 07:36:10
【问题描述】:

我遇到了一个奇怪的问题,我真的不知道如何解决。我正在用 C++ 编写代码,它应该在按下键时触发事件。我可以使用GetAsyncKeystate 很好地检测按键并释放,但我无法可靠地将检测到的按键状态转换为 unicode。 (请注意,我使用的是 Qt,但这并不重要)

我的新闻/发布检测:

// This function is called in a loop
void KeyboardApi::Check()
{
    bool stat = false;
    for (int i = 7; i < 255; i++) { // 0 undefined, 3 VK_CANCEL can be ignored, 1 2 4 5 6 mouse keys, these are ignored
        stat = ((GetAsyncKeyState(i) & 0x8000) != 0);

        if (this->first_time) { // To prevent reporting keypresses upon initialization
            this->first_time = false;
            this->keystate[i] = stat;
            continue;
        }

        if (stat != this->keystate[i]) {
            this->keystate[i] = stat;

            if (i == VK_SHIFT)
                this->shift = stat;
            else if (i == VK_CONTROL)
                this->ctrl = stat;
            else if (i == VK_MENU)
                this->alt = stat;

            HKL locale = GetKeyboardLayout(GetCurrentThreadId());

            // ---!! If this portion is commented, the code does not work correctly.
            //       if this portion is *not* commented, the code works fine...
            /*
            std::wcout << "args for keycode_to_unicode:" << std::endl
                       << "  keycode: " << i << std::endl
                       << "  locale: " << locale << std::endl
                       << "  shift: " << this->shift << std::endl;
            */
            // ---!!

            QString key_string = keycode_to_unicode(i, locale, this->shift);

            if (stat)
                std::wcout << "Key pressed:  " << i <<  " unicode: " << key_string.toStdWString() << std::endl;
        }
    }
}

还有 keycode_to_unicode 函数:

QString keycode_to_unicode(unsigned int key, HKL keyboardLayoutHandle, bool shiftPressed)
{
    int scanCodeEx = MapVirtualKeyExW(key, MAPVK_VK_TO_VSC_EX, keyboardLayoutHandle);

    if (scanCodeEx > 0) {
        unsigned char lpKeyState[256];

        if (shiftPressed) {
            lpKeyState[VK_SHIFT] = 0x80;
            lpKeyState[VK_LSHIFT] = 0x80;
        }

        wchar_t buffer[5];

        int rc = ToUnicodeEx(key, scanCodeEx, lpKeyState, buffer, 5, 0, keyboardLayoutHandle);

        if (rc > 0) {
            return QString::fromWCharArray(buffer);
        } else {
            // It's a dead key; let's flush out whats stored in the keyboard state.
            rc = ToUnicodeEx(key, scanCodeEx, lpKeyState, buffer, 5, 0, keyboardLayoutHandle);
            return QString();
        }
    }

    return QString();
}

所以奇怪的是,当我在那里有调试输出时,KeyboardApi::Check() 工作正常,但是当我没有它时,转换为 unicode 会出错。例如,当我第一次按下'A'键时,输出'a'。第二次,'β','β'等

编辑: 你可能会问自己为什么我不使用 Qt 的内置 onKeyPress... 这是因为我的代码被注入到其他进程中,所以我不能使用这种方法。

【问题讨论】:

    标签: c++ winapi keypress


    【解决方案1】:

    原来是keycode_to_unicode的问题!

    在我的代码中,我有:

    unsigned char lpKeyState[256];
    

    表示数组未初始化;它的内容可以是任何东西。替换为:

    unsigned char lpKeyState[256] = {0};
    

    修复了奇数输入的问题。原来我还有一些其他问题(特别是死键)。我稍微修改了我的代码,现在能够在按下死键时正确检测到死键,尽管我还没有成功地将死键与后续字符组合以检测带重音的字符。为了完整起见,下面是修改后的代码:

    注意:它确实考虑大写锁定。

    QString KeyboardApi::keycode_to_unicode(unsigned int key, HKL keyboardLayoutHandle, bool shiftPressed, bool ctrlPressed, bool altPressed, bool capslock)
    {
        int scanCodeEx = MapVirtualKeyExW(key, MAPVK_VK_TO_VSC_EX, keyboardLayoutHandle);
    
        if (scanCodeEx <= 0)
            return QString();
    
        unsigned char lpKeyState[256] = {0};
    
        if (shiftPressed) {
            lpKeyState[VK_SHIFT] = 0x80;
            lpKeyState[VK_LSHIFT] = 0x80;
        }
    
        if (ctrlPressed)
            lpKeyState[VK_CONTROL] = 0x80;
    
        if (altPressed)
            lpKeyState[VK_MENU] = 0x80;
    
        if (capslock)
            lpKeyState[VK_CAPITAL] = 0x01;
    
        wchar_t buffer[5];
    
        ToUnicodeEx(key, scanCodeEx, lpKeyState, buffer, 5, 0, keyboardLayoutHandle);
    
        QString ret = QString::fromWCharArray(buffer);
        ret.truncate(1);
    
        return ret;
    }
    
    Qt::Key KeyboardApi::unicode_to_qtkey(QString key)
    {
        unsigned int code = key[0].unicode();
        if (key[0].unicode() >= 'a' && key[0].unicode() <= 'z') {
            code = key[0].unicode() - 'a' + Qt::Key_A;
        }
    
        return (Qt::Key)code;
    }
    
    void KeyboardApi::Check()
    {
        bool init = this->first_time;
    
        if (this->first_time)
            this->first_time = false;
    
        bool stat = false;
        for (int i = 7; i < 255; i++) { // yes, 255 is NOT valid! // 0 undefined, 3 VK_CANCEL can be ignored, 1 2 4 5 6 mouse keys
            stat = ((GetAsyncKeyState(i) & 0x8000) != 0);
    
            if (stat == this->keystate[i])
                continue;
    
            this->keystate[i] = stat;
    
            if (i == VK_SHIFT)
                this->shift = stat;
            else if (i == VK_CONTROL)
                this->ctrl = stat;
            else if (i == VK_MENU)
                this->alt = stat;
            else if (i == VK_CAPITAL)
                this->capslock = stat;
    
            HKL locale = GetKeyboardLayout(GetCurrentThreadId());
    
            QString key_string = keycode_to_unicode(i, locale, this->shift, this->ctrl, this->alt, false);
    
            // See qkeymapper_win.cpp for the table used to convert windows virtual key codes to Qt::Key
            Qt::Key key = key_string.isEmpty() ? (Qt::Key)win_vk_to_qt_key[i] : unicode_to_qtkey(key_string);
    
            if (init || i == VK_SHIFT || i == VK_CONTROL || i == VK_MENU)
                continue;
    
            if (stat) {
                emit this->OnKeyDown(this, QKeyEvent(QEvent::KeyPress, key, mods, key_string));
            } else
                emit this->OnKeyUp(this, QKeyEvent(QEvent::KeyPress, key, mods, key_string));
        }
    }
    

    【讨论】:

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