求助，寫的單片機按鍵矩陣不能實現功能

ID:376790 · 發表于 2020-1-17 16:34

寫的是用定時器掃描矩陣按鍵，然后在數碼管顯示，可是代碼不能實現功能，檢查了幾遍也沒想明白哪里出錯了，單片機小白，求助

#include "reg52.h"
#include "intrins.h"
void anjian();
unsigned int a,b;
sbit KEY_IN1 = P3^7;
sbit KEY_IN2 = P3^6;
sbit KEY_IN3 = P3^5;
sbit KEY_IN4 = P3^4;
sbit KEY_OUT1 = P3^0;
sbit KEY_OUT2 = P3^1;
sbit KEY_OUT3 = P3^2;
sbit KEY_OUT4 = P3^3;
unsigned char code pucseg_code[] = {
0xc0, 0xf9, 0xa4, 0xb0, 0x99, 0x92, 0x82, 0xf8,
0x80, 0x90, 0x88, 0x83, 0xc6, 0xa1, 0x86, 0x8e
}; //數碼管顯示段碼
/*unsigned char ledbuff[8] = {
0xff, 0xff, 0xff, 0xff, 0xff, 0xff, 0xff, 0xff
};*/
unsigned char keysta[4][4] = {
{1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1}
}; //矩陣按鍵當前狀態
void Delay1ms() //@12.000MHz
{
unsigned char i, j;
i = 12;
j = 169;
do
{
while (--j);
} while (--i);
}
void Delay300us() //@12.000MHz
{
unsigned char i, j;
i = 4;
j = 125;
do
{
while (--j);
} while (--i);
}
void cls_periphercal(void) //系統初始化
{
P0 = 0xff;
P2 = P2 & 0x1f | 0x80;
P2 = 0x1f;
P0 = 0;
P2 = P2 & 0x1f | 0xa0;
P2 = 0x1f;
}
void seg_disp(unsigned char pu, unsigned char ucseg_pos) //數碼管段選
{
P0 = 0xff;
P2 = P2 & 0x1f | 0xe0;
P2 &= 0x1f;
Delay300us();
P0 = 0x80 >> ucseg_pos;
P2 = P2 & 0x1f | 0xc0;
P2 &= 0x1f;
Delay300us();
P0 = pu;
P2 = P2 & 0x1f | 0xe0;
P2 &= 0x1f;
Delay300us();
}
void saomiao(unsigned char i) //數碼管循環刷新顯示
{
switch(i)
{
case 0: seg_disp(pucseg_code[0], 0); break;
case 1: seg_disp(pucseg_code[1], 1); break;
case 2: seg_disp(pucseg_code[2], 2); break;
case 3: seg_disp(pucseg_code[3], 3); break;
case 4: seg_disp(pucseg_code[4], 4); break;
case 5: seg_disp(pucseg_code[5], 5); break;
case 6: seg_disp(pucseg_code[6], 6); break;
case 7: seg_disp(pucseg_code[7], 7); break;
}
}
/*void Timer0Init(void) //2??@12.000MHz //定時器定時2ms
{
AUXR |= 0x80; //?????1T??
TMOD &= 0xF0; //???????
TL0 = 0x40; //??????
TH0 = 0xA2; //??????
TF0 = 0; //??TF0??
TR0 = 1; //???0????
ET0 = 1;
EA = 1;
}*/
void InterruptTimer0() interrupt 1
{
unsigned char i;
static unsigned char keyout = 0; //輸出索引值
static unsigned char keybuf[4][4] = { //掃描緩沖數組
{0xff, 0xff, 0xff, 0xff}, {0xff, 0xff, 0xff, 0xff},
{0xff, 0xff, 0xff, 0xff}, {0xff, 0xff, 0xff, 0xff}
};
TH0 = 0XFC;
TL0 = 0X67;
keybuf[keyout][0] = (keybuf[keyout][0] << 1) | P3^7;
keybuf[keyout][0] = (keybuf[keyout][0] << 1) | P3^6;
keybuf[keyout][0] = (keybuf[keyout][0] << 1) | P3^5;
keybuf[keyout][0] = (keybuf[keyout][0] << 1) | P3^4;
for(i=0; i<4; i++)
{
if((keybuf[keyout][i] & 0x0f) == 0x00)
keysta[keyout][i] = 0; //連續掃描四次4值為0，可以認為按鍵已經穩定按下或彈起
else if((keybuf[keyout][i] & 0x0f) == 0x0f)
keysta[keyout][i] = 1;
}
keyout++; //索引值遞增
keyout = keyout & 0x03; //當索引值到4時歸零
switch(keyout) //根據索引值，釋放當前輸出引腳，拉低下次輸出引腳
{
case 0 : KEY_OUT4 = 1; KEY_OUT1 = 0; break;
case 1 : KEY_OUT1 = 1; KEY_OUT2 = 0; break;
case 2 : KEY_OUT2 = 1; KEY_OUT3 = 0; break;
case 3 : KEY_OUT3 = 1; KEY_OUT4 = 0; break;
default: break;
}
}
void anjian(void) //判斷按鍵按下的是哪一位
{
if(b == 0 & a == 0)
{
saomiao(0);
Delay1ms();
}
else if(b == 1 & a == 0)
{
saomiao(1);
Delay1ms();
}
else if(b == 2 & a == 0)
{
saomiao(2);
Delay1ms();
}
else if(b == 3 & a == 0)
{
saomiao(3);
Delay1ms();
}
else if(b == 0 & a == 1)
{
saomiao(4);
Delay1ms();
}
else if(b == 1 & a == 1)
{
saomiao(5);
Delay1ms();
}
else if(b == 2 & a == 1)
{
saomiao(6);
Delay1ms();
}
else if(b == 3 & a == 1)
{
saomiao(7);
Delay1ms();
}
else if(b == 0 & a == 2)
{
saomiao(8);
Delay1ms();
}
else if(b == 1 & a == 2)
{
saomiao(9);
Delay1ms();
}
}
void main(void)
{
unsigned char backup[4][4] = {
{1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1}, {1, 1, 1, 1}
}; //按鍵掃描值備份
EA = 1;
//ENLED = 0;
TMOD = 0X01;
TH0 = 0XFC;
TL0 = 0X67;
ET0 = 1;
TR0 = 1;
saomiao(0);
cls_periphercal();
Timer0Init();
while(1)
{
for(a=0; a<4; a++)
{
for(b=0; b<4; b++)
{
if(backup[a][b] != keysta[a][b])
{
if(backup[a][b] != 0)
{
//anjian();
saomiao(7);
Delay1ms();
Delay1ms();
Delay1ms();
Delay1ms();
Delay1ms();
Delay1ms();
}
backup[a][b] = keysta[a][b];
}
}
}
}
}

復制代碼

ID:376790 · 發表于 2020-1-17 16:44

求助

ID:473159 · 發表于 2020-1-17 20:08

一個一個按鍵的調試，屏蔽其他按鍵，先調通一個，后面就會照做就OK!

ID:213173 · 發表于 2020-1-17 22:12

sengsoudang 發表于 2020-1-17 16:44
求助

樓主這是從哪里東拼西湊來的代碼，只是用數碼管顯示矩陣按鍵鍵值的簡單程序，竟然洋洋灑灑幾百條代碼。給你簡化一下經仿真無誤。無標題.jpg

#include "reg52.h"
#include "intrins.h"
#define uint unsigned int
#define uchar unsigned char
uchar code pucseg_code[] = {//段碼
0xc0, 0xf9, 0xa4, 0xb0, 0x99, 0x92, 0x82, 0xf8,
0x80, 0x90, 0x88, 0x83, 0xc6, 0xa1, 0x86, 0x8e};
uchar code weia[] = {0x80,0x40,0x20,0x10,0x08,0x04,0x02,0x01};//位碼
uchar data dis_buf[8];//顯示緩存
uchar key=0; //定義鍵值全局變量
void anjian() //按鍵掃描程序
{
static bit sign=0; //按鍵有效標志
static uchar count=0; //消抖計數變量
uchar num=0; //臨時變量
P3=0xf0; //賦值P3 1111 0000
if(P3!=0xf0) //檢測有按鍵按下
{
count++; //消抖計數
if((count>=10)&&(sign==0))
{
sign=1; //按鍵有效標志置1
num=P3; //保存P3值xxxx 0000，x為0或1
num|=0x0f; //保存num按位或0x0f值xxxx 1111
P3=num; //賦值P3 xxxx 1111
switch(P3)
{
case 0xee: key= 1; break;
case 0xde: key= 2; break;
case 0xbe: key= 3; break;
case 0x7e: key= 4; break;
case 0xed: key= 5; break;
case 0xdd: key= 6; break;
case 0xbd: key= 7; break;
case 0x7d: key= 8; break;
case 0xeb: key= 9; break;
case 0xdb: key=10; break;
case 0xbb: key=11; break;
case 0x7b: key=12; break;
case 0xe7: key=13; break;
case 0xd7: key=14; break;
case 0xb7: key=15; break;
case 0x77: key=16; break;
}
}
}
else //鍵抬起
{
sign=0; //按鍵有效標志清0
count=0; //消抖計數清0
}
}
void saomiao()//共陽數碼管循環刷新顯示
{
static bit i=0;
if(i)
dis_buf[0]=pucseg_code[key%10];
else
dis_buf[1]=pucseg_code[key/10];
P0=0xff;//消隱
P2=weia[i];
P0=dis_buf[i];
i=~i;
}
void main() //主函數
{
TMOD = 0X01;//1ms
TH0 = 0XFC;
TL0 = 0X67;
EA = 1;
ET0 = 1;
TR0 = 1;
while(1);
}
void InterruptTimer0() interrupt 1
{
TH0 = 0XFC;
TL0 = 0X67;
anjian(); //按鍵掃描程序
saomiao(); //共陽數碼管顯示程序
}

復制代碼

ID:376790 · 發表于 2020-1-18 14:22

wulin 發表于 2020-1-17 22:12
樓主這是從哪里東拼西湊來的代碼，只是用數碼管顯示矩陣按鍵鍵值的簡單程序，竟然洋洋灑灑幾百條代碼。給 ...

自己參考書里改的

，感謝大佬，我參考

ID:376790 · 發表于 2020-1-18 14:23

yoxi 發表于 2020-1-17 20:08
一個一個按鍵的調試，屏蔽其他按鍵，先調通一個，后面就會照做就OK!

好的，感謝解答

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求助，寫的單片機按鍵矩陣不能實現功能