这是一个4行6列矩阵按键原代码
- #include <reg52.h>
- sbit KEY_IN_0 = P1^0;
- sbit KEY_IN_1 = P1^1;
- sbit KEY_IN_2 = P1^2;
- sbit KEY_IN_3 = P1^3;
- sbit KEY_IN_4 = P1^4;
- sbit KEY_IN_5 = P1^5;
- sbit KEY_OUT_1 = P3^0;
- sbit KEY_OUT_2 = P3^1;
- sbit KEY_OUT_3 = P3^2;
- sbit KEY_OUT_4 = P3^6;
- unsigned char code KeyCodeMap[4][6] = { //矩阵按键编号到标准键盘键码的映射表
- { 0x1B,'1', '2', '3', 0x2A, 0x27 }, //ESC 键、数字键 1、数字键 2、数字键 3、向上键
- { 0x1A,'4', '5', '6', 0x2B, 0x26 }, //开关屏、数字键 4、数字键 5、数字键 6、向左键
- { 0x1C,'7', '8', '9', 0x2D, 0x25 }, //电机开、数字键 7、数字键 8、数字键 9、向下键
- { 0x1D,0X06,'0',0x2F, 0x0D, 0x28 } //功能键、DEL 键、 数字键 0、回车键、 向右键
- };
- unsigned char pdata KeySta[4][6] = { //全部矩阵按键的当前状态
- {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}
- };
- extern void KeyAction(unsigned char keycode);
- /* 按键驱动函数,检测按键动作,调度相应动作函数,需在主循环中调用 */
- void KeyDriver()
- {
- unsigned char i, j;
- static unsigned char pdata backup[4][6] = { //按键值备份,保存前一次的值
- {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}
- };
- for (i=0; i<4; i++) //循环检测 4*4 的矩阵按键
- {
- for (j=0; j<6; j++)
- {
- if (backup[i][j] != KeySta[i][j]) //检测按键动作
- {
- if (backup[i][j] != 0) //按键按下时执行动作
- {
- KeyAction(KeyCodeMap[i][j]); //调用按键动作函数
- }
- backup[i][j] = KeySta[i][j]; //刷新前一次的备份值
- }
- }
- }
- }
- /* 按键扫描函数,需在定时中断中调用,推荐调用间隔 1ms */
- void KeyScan()
- {
- unsigned char i;
- static unsigned char keyout = 0; //矩阵按键扫描输出索引
- static unsigned char keybuf[4][6] = { //矩阵按键扫描缓冲区
- {0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF}, {0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF},
- {0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF}, {0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF}
- };
- //将一行的 4 个按键值移入缓冲区
- keybuf[keyout][0] = (keybuf[keyout][0] << 1) | KEY_IN_0;
- keybuf[keyout][1] = (keybuf[keyout][1] << 1) | KEY_IN_1;
- keybuf[keyout][2] = (keybuf[keyout][2] << 1) | KEY_IN_2;
- keybuf[keyout][3] = (keybuf[keyout][3] << 1) | KEY_IN_3;
- keybuf[keyout][4] = (keybuf[keyout][4] << 1) | KEY_IN_4;
- keybuf[keyout][5] = (keybuf[keyout][5] << 1) | KEY_IN_5;
- //消抖后更新按键状态
- for (i=0; i<6; i++) //每行 4 个按键,所以循环 4 次
- {
- if ((keybuf[keyout][i] & 0x0F) == 0x00)
- { //连续 4 次扫描值为 0,即 4*4ms 内都是按下状态时,可认为按键已稳定的按下
- KeySta[keyout][i] = 0;
- }
- else if ((keybuf[keyout][i] & 0x0F) == 0x0F)
- { //连续 4 次扫描值为 1,即 4*4ms 内都是弹起状态时,可认为按键已稳定的弹起
- KeySta[keyout][i] = 1;
- }
- }
- //执行下一次的扫描输出
- keyout++; //输出索引递增
- keyout &= 0x03; //索引值加到 4 即归零
- switch (keyout) //根据索引,释放当前输出引脚,拉低下次的输出引脚
- {
- case 0: KEY_OUT_4 = 1; KEY_OUT_1 = 0; break;
- case 1: KEY_OUT_1 = 1; KEY_OUT_2 = 0; break;
- case 2: KEY_OUT_2 = 1; KEY_OUT_3 = 0; break;
- case 3: KEY_OUT_3 = 1; KEY_OUT_4 = 0; break;
- default: break;
- }
- }
复制代码
|