51单片机一个按键控制八个灯,四种形态的流水灯 另一个按键控制暂停和继续
时间: 2023-05-15 08:07:01 浏览: 104
这个问题可以通过编写一个简单的程序来实现。以下是一个可能的解决方案:
```c
#include <reg51.h>
// 定义控制灯的端口
sbit LED1 = P1^0;
sbit LED2 = P1^1;
sbit LED3 = P1^2;
sbit LED4 = P1^3;
sbit LED5 = P1^4;
sbit LED6 = P1^5;
sbit LED7 = P1^6;
sbit LED8 = P1^7;
// 定义控制按键的端口
sbit KEY1 = P2^0;
sbit KEY2 = P2^1;
// 定义流水灯的四种形态
unsigned char pattern1[] = {0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40, 0x80};
unsigned char pattern2[] = {0x81, 0x42, 0x24, 0x18, 0x18, 0x24, 0x42, 0x81};
unsigned char pattern3[] = {0x81, 0xC3, 0xE7, 0xFF, 0xFF, 0xE7, 0xC3, 0x81};
unsigned char pattern4[] = {0x81, 0x42, 0x24, 0x18, 0x18, 0x24, 0x42, 0x81};
// 定义流水灯的当前形态和状态
unsigned char* pattern = pattern1;
unsigned char paused = 0;
void main() {
// 初始化按键状态
KEY1 = 1;
KEY2 = 1;
while (1) {
// 检测按键1是否按下
if (KEY1 == 0) {
// 切换流水灯的形态
if (pattern == pattern1) {
pattern = pattern2;
} else if (pattern == pattern2) {
pattern = pattern3;
} else if (pattern == pattern3) {
pattern = pattern4;
} else if (pattern == pattern4) {
pattern = pattern1;
}
// 等待按键1释放
while (KEY1 == 0);
}
// 检测按键2是否按下
if (KEY2 == 0) {
// 切换流水灯的暂停状态
paused = !paused;
// 等待按键2释放
while (KEY2 == 0);
}
// 如果没有暂停,则显示流水灯的当前形态
if (!paused) {
LED1 = pattern[0];
LED2 = pattern[1];
LED3 = pattern[2];
LED4 = pattern[3];
LED5 = pattern[4];
LED6 = pattern[5];
LED7 = pattern[6];
LED8 = pattern[7];
}
}
}
```
这个程序使用了一个无限循环来检测按键的状态,并根据按键的状态来切换流水灯的形态和暂停状态。在每次循环中,程序会检测按键1和按键2是否按下,如果按下则执行相应的操作。如果没有暂停,则程序会根据当前的流水灯形态来控制灯的状态。