蓝桥杯单片机必备知识—–(6)矩阵按键

思路：就是线反转法

线反转法：将控制行的线置高控制列的线置地，如果哪个低，则为哪一行有按键按下；然后将控制列的线置高控制行的线置地，如果哪个低，则为哪一列有按键按下。如此之后就可以确定到底是哪个按键按下。

注：跳线帽设置为KBD！！！！！

扫描键盘函数

void read_key()
{
   
	static unsigned char hang;
	static unsigned char state = 0;
	
	switch(state)
	{
   
		case 0:	
		{
   
			P3 = 0x0f;P42 = 0;P44 = 0; 
			if(P3 != 0x0f) state = 1;
		}break;
		case 1:
		{
   
			P3 = 0x0f; P42 = 0; P44 = 0;	//列置低，行置高
			if(P3 != 0x0f) 		//判断行
			{
   
				if(P33 == 0) hang = 1;
				if(P32 == 0) hang = 2;
				if(P31 == 0) hang = 3;
				if(P30 == 0) hang = 4;
				switch(hang)
				{
   
					case 1:
					{
   
						P3 = 0xf0;P42 = 1;P44 = 1;				//行置低，列置高
						if(P44 == 0){
    s4++;state = 2;hang = 0;}
						if(P42 == 0){
    s8++;state = 2;hang = 0;}
						if(P35 == 0){
   s12++;state = 2;hang = 0;}
						if(P34 == 0){
    s16++;state = 2;hang = 0;}
					}break;
					case 2:
					{
   
						P3 = 0xf0;P42 = 1;P44 = 1;
						if(P44 == 0){
    s5++;state = 2;hang = 0;}
						if(P42 == 0){
    s9++;state = 2;hang = 0;}
						if(P35 == 0){
   s13++;state = 2;hang = 0;}
						if(P34 == 0){
    s17++;state = 2;hang = 0;}
					}break;
					case 3:
					{
   
						P3 = 0xf0;P42 = 1;P44 = 1;
						if(P44 == 0){
    s6++;state = 2;hang = 0;}
						if(P42 == 0){
    s10++;state = 2;hang = 0;}
						if(P35 == 0){
   s14++;state = 2;hang = 0;}
						if(P34 == 0){
    s18++;state = 2;hang = 0;}
					}break;
					case 4:
					{
   
						P3 = 0xf0;P42 = 1;P44 = 1;
						if(P44 == 0){
    s7++;state = 2;hang = 0;}
						if(P42 == 0){
    s11++;state = 2;hang = 0;}
						if(P35 == 0){
   s15++;state = 2;hang = 0;}
						if(P34 == 0){
    s19++;state = 2;hang = 0;}
					}break;
				}
			}
			else state = 0;
		}break;
		case 2:		//判断按键是否松开
		{
   
			P3 = 0x0f;P42 = 0;P44 = 0;
			if(P3 == 0x0f) state = 0;
		}break;
	}
	
}

测试效果图：

一般蓝桥杯会使用s4s5s8s9小矩形键盘来测试你对矩阵键盘的掌握程度。

小矩阵键盘

void read_key()
{
   
	static unsigned char hang;
	static unsigned char state = 0;
	
	switch(state)
	{
   
		case 0:	
		{
   
			P3 = 0x0f;P42 = 0;P44 = 0;
			if(P3 != 0x0f) state = 1;
		}break;
		case 1:
		{
   
			P3 = 0x0f; P42 = 0; P44 = 0;
			if(P3 != 0x0f) 
			{
   
				if(P33 == 0) hang = 1;
				if(P32 == 0) hang = 2;
				switch(hang)
				{
   
					case 1:
					{
   
						P3 = 0xf0;P42 = 1;P44 = 1;
						if(P44 == 0){
    s4++;state = 2;hang = 0;}
						if(P42 == 0){
    s8++;state = 2;hang = 0;}
					}break;
					case 2:
					{
   
						P3 = 0xf0;P42 = 1;P44 = 1;
						if(P44 == 0){
    s5++;state = 2;hang = 0;}
						if(P42 == 0){
    s9++;state = 2;hang = 0;}
					}break;
				}
			}
			else state = 0;
		}break;
		case 2:
		{
   
			P3 = 0x0f;P42 = 0;P44 = 0;
			if(P3 == 0x0f) state = 0;
		}break;
	}
	
}