这里关闭所有的中断for(inti=0;i<(c<<1);i++){switch_led();delay(switch_interval);}long_press=false;//关灯,清除状态digitalWrite(LED,LOW);interrupts();// 恢复中断事件的响应}voidloop(){do{if(!button_acted||(current-last_switch
However, you want the program to react to a button press to start another activity. While polling the serial port, your program does nothing else and this is where interrupts come in. [Yes I know the serial port has its own interrupt - give me a break - its just an example for ...
(ledPin, OUTPUT); // 设置LED引脚为输出 pinMode(buttonPin, INPUT_PULLUP); // 设置按钮引脚为输入,并启用内部上拉电阻 attachInterrupt(digitalPinToInterrupt(buttonPin), doButtonPress, FALLING); // 设置外部中断,当按钮按下时触发 } void loop() { // 主循环不需要做任何事情,因为中断服务程序会处理...
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void setupInterrupt(){ timer = timerBegin(0, 80, true); // 使用定时器0，预分频器为80，计数...
RDA5807M radio; // Create an instance of Class for RDA5807M ChipRADIO_INFO ri;//rssi 场强检测,用于静噪float fm_freq = 108;void rx(void);void tx(void);void setup() { pinMode(3, INPUT); //sets interrupt to check for button talk abutton press 端口3和接地之间接一个10K电阻,当此端...
Use Interrupts to detect the button press. The first option is easier and is explained morehere. The second is more advanced and involves using low level interrupt routines and more hardware (to allow several keys to operate into one interrupt pin). ...
Turn on the LED when button is pressed, turn it off otherwise What we want to achieve is simple: when the button is not pressed, the LED is off. And when we press the button the LED should be on. The code #define LED_PIN 8 ...
Toggle LED’s state with the push button – first iterationWhat we want to do is to toggle the LED’s state when you press + release the button. So, the first time you release the button, the LED will turn on. The second time, it will turn off. Etc....
搜索研究了一下,好像是因为库里用了一堆interrupt和delay函数(啊这…)。我没敢自己修改库文件,于是用和计数器防止重复类似的手段给LCD也增加了两个函数,只有在计数器数字和之前已经输出的数字不同的情况下才刷新LCD,这样舵机平时就不会抖动了,也算是达到了目的。