求大神,51单片机矩阵键盘问题!!!

2025-03-15 18:26:17
推荐回答(4个)
回答(1):

我给个程序你参考一下

程序:

#include 

#include 

#define uchar unsigned char

#define uint unsigned int

uchar key=16,key_t=16;

sbit duan=P2^0;

sbit wei=P2^1;

uchar code table[]={

0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07, //共阴数码管段选码表,无小数点

           

0x7f,0x6f,0x77,0x7c,0x39,0x5e,0x79,0x71,0x00};

void delay(uint ms)

{

uchar i;

while(ms--)

for(i=0;i<110;i++);

}

void key_tst()

{

uchar kess;

if(P1!=0xf0)

 delay(5);

 if(P1!=0xf0)

 {

P1=0x0f;

    delay(3);

kess=P1^0x0f;    //P1口的状态导或上0x0f,如果下了高4位不管,看低4位是多少

switch (kess)

{

case 1:key=0; break;

case 2:key=1; break;

case 4:key=2; break;

case 8:key=3; break;

default: key = 16;

}

   P1=0xf0;

delay(1);

kess=P1>>4^0x0f;

switch (kess)

{

case 1:key+=12;break;

case 2:key+=8; break;

case 4:key+=4; break;

case 8:key+=0; break;

default: key = 16;

}

 }

}

void main()

P0=0x00;

while(1)

{

key_tst();

P0=table[key];

duan=1;

duan=0;

P0=0x7f;

wei=1;

wei=0;

delay(2);

}

}

回答(2):

我记得矩阵键盘不是你这样扫描的吧,得先扫描行(列),然后扫描列(行),综合得到是哪个键按下吧。并不是一个扫描就行的啊。。。。。。。。

回答(3):

这个不调还真不知道,建议把key改为int型试下。
逻辑上应该没错的。

回答(4):

刚刚我把你的程序下载到得我的PROTUSE键盘扫描电路里面仿真了一下,程序没有问题,肯定是你硬件电路的问题。在我这没有出现你所说的问题。如果你是用开发板仿真的估计是键盘的问题吧,如果是用PROTUES仿真的那估计是你的BUTTON有问题!

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