为什么我按下按钮后无法调用此方法？

Question

我使用按钮作为切换开关。按它，它会做“东西A”。再次按下它，它会做“东西B”。为什么我无法打电话给我的方法checkButtons_slow()？

int prev = 0; 
int current = 0; 
int val4 = 0; 
int val5 = 0; 
int ledPin = 13; 
int prev = 0; 
int current = 0; 

Servo ZServo; 

void setup() { 
    ZServo.attach(9); 
    pinMode(pushD3, INPUT_PULLUP); 
    digitalWrite(3, HIGH); 
    pinMode(pushD4, INPUT_PULLUP); 
    digitalWrite(4, HIGH); 
    pinMode(pushD5, INPUT_PULLUP); 
    digitalWrite(5, HIGH); 
    pinMode(pushD6, INPUT_PULLUP); 
    digitalWrite(6, HIGH); 
    pinMode(ledPin, OUTPUT); 
} 

void loop() { 
    if(digitalRead(3) == LOW) { 
    current = 1 - current; 
    } 
    if(current == 1 && prev == 0) { 
    checkButtons_slow(); 
    //test: ZServo.write(110); 
    delay(500); //half a second 
    } 
    if(current == 0 && prev == 1) { 
    ZServo.write(80); 
    delay(500); //half a second 
    } 
    prev = current; 
} 

这里是我的方法：

void checkButtons_slow() { 
    val4 = digitalRead(pushD4); 
    val5 = digitalRead(pushD5); 
    if (val4 == LOW) { 
    ZServo.write(88); 
    } else if (val5 == LOW) { 
    ZServo.write(99); 
    } else { 
    ZServo.write(91); //GUESSED ON 92; SHOULD TECHNICALLY BE 90 
    } 
} 

所以注释掉//test: ZServo.write(110);作品。我错过了什么checkButtons_slow();？

Answer 1

如果您将void loop()更改为此，则可以开启和关闭方法。

void loop() { 
    if (digitalRead(3) == LOW) { 
    num_presses++; 
    delay(500); 
    } 
    if ((num_presses % 2) == 0) { 
    //even 
    checkButtons_slow(); 
    } 
    else if(num_presses == 0) { 
    ZServo.write(90); 
    } 
    else { 
    ZServo.write(85); 
    } 
}