用独立按键控制LED灯和数码管和蜂鸣器写一个代码

当我们要用独立按键控制LED灯、数码管和蜂鸣器时，通常需要先设置硬件连接，然后编写相应的控制程序。这里以Python示例给出基本的控制流程：

import RPi.GPIO as GPIO  # 使用RPi.GPIO库处理GPIO操作

# 定义GPIO引脚
led_pin = 18
数码管_pin = 21
buzzer_pin = 26

# 设置GPIO模式为BCM
GPIO.setmode(GPIO.BCM)

# 对于LED灯，将其设置为输出
GPIO.setup(led_pin, GPIO.OUT)
led_state = False  # 初始状态为关闭

# 对于数码管和蜂鸣器，假设它们有多个段/键位，并已初始化
digit_pins = [24, 23, 25]  # 数码管各段引脚
digit_buffer = [' ', '0', '1', '2', '3', '4', '5', '6', '7', '8', '9']  # 指定显示字符

def toggle_led():
    global led_state
    led_state = not led_state
    GPIO.output(led_pin, led_state)

def update_digit(num):
    digit_pins_str = ''.join([digit_pins[i] for i in range(len(digit_pins)) if num & (1 << i)])
    GPIO.output(digit_pins_str, GPIO.HIGH)  # 高电平表示点亮相应数码管段
    GPIO.output([pin for pin in digit_pins if pin not in digit_pins_str], GPIO.LOW)  # 熄灭其他段

def play_buzzer(duration=1):  # 假设duration是以秒为单位的音长
    GPIO.setup(buzzer_pin, GPIO.OUT)
    GPIO.output(buzzer_pin, GPIO.HIGH)
    time.sleep(duration)
    GPIO.output(buzzer_pin, GPIO.LOW)

# 示例按键事件回调函数
def button_event(channel):
    if channel == 2:  # 假设按钮A控制LED灯
        toggle_led()
    elif channel == 3:  # 假设按钮B控制数码管显示
        num = GPIO.input(buttonB_pin)  # 获取按钮B输入
        update_digit(num)
    elif channel == 4:  # 假设按钮C控制蜂鸣器
        play_buzzer()

# 添加按键中断并注册回调函数
buttonA_pin = 27  # 假设按钮A接在GPIO 27
buttonB_pin = 17  # 假设按钮B接在GPIO 17
buttonC_pin = 12  # 假设按钮C接在GPIO 12
GPIO.add_event_detect(buttonA_pin, GPIO.RISING, callback=button_event, bouncetime=200)
GPIO.add_event_detect(buttonB_pin, GPIO.RISING, callback=button_event, bouncetime=200)
GPIO.add_event_detect(buttonC_pin, GPIO.RISING, callback=button_event, bouncetime=200)

try:
    while True:
        pass  # 循环等待按键触发
except KeyboardInterrupt:
    GPIO.cleanup()  # 键盘中断时清理GPIO资源