帮我写一个按键按下一次，控制亮灯0.6秒，灯灭0.6秒，闪灯循环两次后舵机从0°旋转到90°的程序

好的，下面是程序：

import RPi.GPIO as GPIO
import time

led_pin = 18
servo_pin = 23

GPIO.setmode(GPIO.BCM)
GPIO.setup(led_pin, GPIO.OUT)
GPIO.setup(servo_pin, GPIO.OUT)

def flash_led():
    for i in range(2):
        GPIO.output(led_pin, GPIO.HIGH)
        time.sleep(0.6)
        GPIO.output(led_pin, GPIO.LOW)
        time.sleep(0.6)

def rotate_servo():
    p = GPIO.PWM(servo_pin, 50)
    p.start(0)
    p.ChangeDutyCycle(7.5)
    time.sleep(1)
    p.ChangeDutyCycle(12.5)
    time.sleep(1)
    p.stop()

while True:
    input_state = GPIO.input(4)
    if input_state == False:
        flash_led()
        rotate_servo()

当你按下按键时，程序会循环闪烁LED两次，然后舵机会从0°旋转到90°。注意，此程序需要安装RPi.GPIO库。

相关问题

写一个按键按下一次，控制亮灯0.6秒，灯灭0.6秒，闪灯循环两次后舵机从0°旋转到90°的程序

以下为代码实现：

#include <Arduino.h>

const int buttonPin = 2;
const int ledPin = 13;
const int servoPin = 9;

int buttonState = 0;

void setup() {
  pinMode(buttonPin, INPUT);
  pinMode(ledPin, OUTPUT);
  pinMode(servoPin, OUTPUT);
}

void loop() {
  buttonState = digitalRead(buttonPin);

  if (buttonState == HIGH) {
    digitalWrite(ledPin, HIGH);
    delay(600);
    digitalWrite(ledPin, LOW);
    delay(600);
    digitalWrite(ledPin, HIGH);
    delay(600);
    digitalWrite(ledPin, LOW);
    delay(600);
    digitalWrite(ledPin, HIGH);
    delay(600);
    digitalWrite(ledPin, LOW);
    delay(600);
    digitalWrite(ledPin, HIGH);
    delay(600);
    digitalWrite(ledPin, LOW);

    for (int i = 0; i <= 90; i++) {
      digitalWrite(servoPin, HIGH);
      delay(15);
      digitalWrite(servoPin, LOW);
      delayMicroseconds(1500 - i*10);
    }

    delay(1000);

    for (int i = 90; i >= 0; i--) {
      digitalWrite(servoPin, HIGH);
      delay(15);
      digitalWrite(servoPin, LOW);
      delayMicroseconds(1500 - i*10);
    }
  }
}

当你按下按键时，LED灯会闪烁两次，之后舵机会从0°旋转到90°，再从90°旋转回0°。