Simple keypad library, for a 3-column x 4-row keypad.
Key arrangement is: ---------
1 2 3
4 5 6
7 8 9
* 0 #
---------
Supports two function keys ('*' and '#').
………………
static char Last_valid_key_G = KEYPAD_NO_NEW_DATA;//静态全局变量
………………
void KEYPAD_Update(void) //键盘更新函数
{
char Key, FnKey;
// Scan keypad here...
if (KEYPAD_Scan(&Key, &FnKey) == 0) //调用键盘扫描
{
// No new key data - just return
return;
}
………………
………………
bit KEYPAD_Scan(char* const pKey, char* const pFuncKey)
{
static data char Old_Key;
char Key = KEYPAD_NO_NEW_DATA;
char Fn_key = (char) 0x00;
C1 = 0; // Scanning column 1
if (R1 == 0) Key = '1';
if (R2 == 0) Key = '4';
if (R3 == 0) Key = '7';
if (R4 == 0) Fn_key = '*';
C1 = 1;
C2 = 0; // Scanning column 2
if (R1 == 0) Key = '2';
if (R2 == 0) Key = '5';
if (R3 == 0) Key = '8';
if (R4 == 0) Key = '0';
C2 = 1;
C3 = 0; // Scanning column 3
if (R1 == 0) Key = '3';
if (R2 == 0) Key = '6';
if (R3 == 0) Key = '9';
if (R4 == 0) Fn_key = '#';
C3 = 1;
if (Key == KEYPAD_NO_NEW_DATA)
{
// No key pressed (or just a function key)
Old_Key = KEYPAD_NO_NEW_DATA;
Last_valid_key_G = KEYPAD_NO_NEW_DATA;
return 0; // No new data
}
// A key has been pressed: debounce by checking twice
if (Key == Old_Key) //这步！初始化时
{ //Key==Old_Key是理应的
//当有KEY按下后
//Key！=Old_Key
//但！=Old_Key又如何
//进入IF语句？？？
// A valid (debounced) key press has been detected
//这里小了延时语句？？
// Must be a new key to be valid - no 'auto repeat'
if (Key != Last_valid_key_G)
{
// New key!
*pKey = Key;
Last_valid_key_G = Key;
// Is the function key pressed too?
if (Fn_key)
{
// Function key *is* pressed with another key
*pFuncKey = Fn_key;
}
else
{
*pFuncKey = (char) 0x00;
}
return 1;
}
}
// No new data
Old_Key = Key; //当某KEY被一直按下时，由于不支持重
return 0; //复，虽然检测到Key ！= KEYPAD_NO_NEW_DATA
} //但由于Key == Old_Key，仍然进入IF语句，再由
//Key != Last_valid_key_G 决定不是新键，
//return 0;
个人认为，第一次有KEY按下后，要求Key ！= Old_Key进入IF语句：
但在长按情况下，却要求Key == Old_Key进入IF语句。。。如何解释？？？
|
楼主
标题置顶
标题高亮
