C语言写一个4*16的矩阵开关
时间: 2024-02-01 09:14:27 浏览: 115
4个16*16点阵屏C#上位机代码
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这里给出一个使用单片机的C语言程序,实现4*16的矩阵开关:
```c
#include <reg52.h>
#define ROWS 4
#define COLS 16
// 定义矩阵开关的IO口
sbit ROW1=P1^0;
sbit ROW2=P1^1;
sbit ROW3=P1^2;
sbit ROW4=P1^3;
sbit COL1=P1^4;
sbit COL2=P1^5;
sbit COL3=P1^6;
sbit COL4=P1^7;
sbit COL5=P2^0;
sbit COL6=P2^1;
sbit COL7=P2^2;
sbit COL8=P2^3;
sbit COL9=P2^4;
sbit COL10=P2^5;
sbit COL11=P2^6;
sbit COL12=P2^7;
// 定义矩阵开关的状态数组
unsigned char matrix[ROWS][COLS];
// 初始化矩阵开关的状态数组
void init_matrix(void)
{
int i, j;
for (i = 0; i < ROWS; i++)
{
for (j = 0; j < COLS; j++)
{
matrix[i][j] = 0;
}
}
}
// 设置矩阵开关某个位置的状态
void set_matrix(unsigned char row, unsigned char col, unsigned char state)
{
matrix[row][col] = state;
}
// 获取矩阵开关某个位置的状态
unsigned char get_matrix(unsigned char row, unsigned char col)
{
return matrix[row][col];
}
// 扫描矩阵开关,并更新状态数组
void scan_matrix(void)
{
int i, j;
for (i = 0; i < ROWS; i++)
{
switch (i)
{
case 0:
ROW1 = 0;
ROW2 = 1;
ROW3 = 1;
ROW4 = 1;
break;
case 1:
ROW1 = 1;
ROW2 = 0;
ROW3 = 1;
ROW4 = 1;
break;
case 2:
ROW1 = 1;
ROW2 = 1;
ROW3 = 0;
ROW4 = 1;
break;
case 3:
ROW1 = 1;
ROW2 = 1;
ROW3 = 1;
ROW4 = 0;
break;
default:
break;
}
for (j = 0; j < COLS; j++)
{
switch (j)
{
case 0:
set_matrix(i, j, COL1);
break;
case 1:
set_matrix(i, j, COL2);
break;
case 2:
set_matrix(i, j, COL3);
break;
case 3:
set_matrix(i, j, COL4);
break;
case 4:
set_matrix(i, j, COL5);
break;
case 5:
set_matrix(i, j, COL6);
break;
case 6:
set_matrix(i, j, COL7);
break;
case 7:
set_matrix(i, j, COL8);
break;
case 8:
set_matrix(i, j, COL9);
break;
case 9:
set_matrix(i, j, COL10);
break;
case 10:
set_matrix(i, j, COL11);
break;
case 11:
set_matrix(i, j, COL12);
break;
default:
break;
}
}
}
}
void main()
{
init_matrix();
while (1)
{
scan_matrix();
// 在这里可以根据需要对矩阵开关的状态数组进行处理
}
}
```
注意,这个程序使用了单片机的IO口来控制矩阵开关,如果需要在其他平台上实现,需要根据具体情况进行修改。
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