這是我最終開始創建鍵盤按鍵映射的方式。它涵蓋了幾乎所有的可能性:不可映射的鍵被存儲爲十六進制字符串,可映射的鍵不表示自己爲單個字符需要手動添加。也許這對未來的某個人可能是有用的。
CreateKeyboardMap()
{
string keystring;
char keybuffer;
for(int i = 0; i < 256; ++i){
if(keybuffer = MapVirtualKey(UINT(i),2)){
keystring += keybuffer;
}
else{
keystring = int_to_hex(i);
}
_keyboardMap.insert(_keyboardMap.end(),pair<UINT,string>(i,keystring));
keystring = "";
}
_keyboardMap[0x1B] = "ESCAPE";
_keyboardMap[0x70] = "F1";
_keyboardMap[0x71] = "F2";
_keyboardMap[0x72] = "F3";
_keyboardMap[0x73] = "F4";
_keyboardMap[0x74] = "F5";
_keyboardMap[0x75] = "F6";
_keyboardMap[0x76] = "F7";
_keyboardMap[0x77] = "F8";
_keyboardMap[0x78] = "F9";
_keyboardMap[0x79] = "F10";
_keyboardMap[0x7A] = "F11";
_keyboardMap[0x7B] = "F12";
_keyboardMap[0x2C] = "PRINT SCREEN";
_keyboardMap[0x91] = "SCROLL LOCK";
_keyboardMap[0x08] = "BACKSPACE";
_keyboardMap[0x20] = "SPACE";
_keyboardMap[0x2D] = "INSERT";
_keyboardMap[0x24] = "HOME";
_keyboardMap[0x22] = "PAGE DOWN";
_keyboardMap[0x21] = "PAGE UP";
_keyboardMap[0x2E] = "DELETE";
_keyboardMap[0x90] = "NUMLOCK";
_keyboardMap[0x6F] = "NUMPAD /";
_keyboardMap[0x6A] = "NUMPAD *";
_keyboardMap[0x6D] = "NUMPAD -";
_keyboardMap[0x6B] = "NUMPAD +";
_keyboardMap[0x6E] = "NUMPAD .";
_keyboardMap[0x60] = "NUMPAD 0";
_keyboardMap[0x61] = "NUMPAD 1";
_keyboardMap[0x62] = "NUMPAD 2";
_keyboardMap[0x63] = "NUMPAD 3";
_keyboardMap[0x64] = "NUMPAD 4";
_keyboardMap[0x65] = "NUMPAD 5";
_keyboardMap[0x66] = "NUMPAD 6";
_keyboardMap[0x67] = "NUMPAD 7";
_keyboardMap[0x68] = "NUMPAD 8";
_keyboardMap[0x68] = "NUMPAD 9";
_keyboardMap[0x26] = "ARROW UP";
_keyboardMap[0x28] = "ARROW DOWN";
_keyboardMap[0x25] = "ARROW LEFT";
_keyboardMap[0x27] = "ARROW RIGHT";
_keyboardMap[0x0D] = "ENTER";
_keyboardMap[0xA0] = "LSHIFT";
_keyboardMap[0xA1] = "RSHIFT";
_keyboardMap[0x09] = "TAB";
_keyboardMap[0x14] = "CAPS LOCK";
_keyboardMap[0xA2] = "LCONTROL";
_keyboardMap[0xA3] = "RCONTROL";
_keyboardMap[0xA4] = "LALT";
_keyboardMap[0xA5] = "RALT";
_keyboardMap[0x5B] = "LWIN";
_keyboardMap[0x5C] = "RWIN";
}
沒錯,但問題是如何遍歷所有可能的密鑰,所有的虛擬鍵有一個十六進制值,並且地圖會在該點創建的條目只對指定的鍵 – Smash
既然你實際上顯示這是一個for循環,我認爲很明顯,這不是真正的問題。 –
for循環是爲了說明問題...因此_有一種方法來迭代所有可能的虛擬鍵_並在最後_或我需要手動檢查每一個?_ – Smash