无效使用 void 表达式，怎么办？

Question

不知道为什么我会收到这个。

该程序编码错误，是在 Arduino Uno 上运行开发的地鼠游戏。

我的 IED 是 Arduino IED 版本 1.6.12

是的，还有很多内容，但选择省略，如果您想查看所有代码，请询问。

错误

Arduino: 1.6.12 (Windows 7), Board: "Arduino/Genuino Uno"

C:\Users\Strings\Documents\Arduino\Too_Many_Moles\Too_Many_Moles.ino: In function 'void winOrLost()':

Too_Many_Moles:146: error: invalid use of void expression

       timelyCall(current_millis, toggleBuzer(), 1000);

                                                     ^

C:\Users\Strings\Documents\Arduino\Too_Many_Moles\Too_Many_Moles.ino: In function 'void loop()':

Too_Many_Moles:191: error: invalid use of void expression

     timelyCall(current_millis, popAMole(), time_interval);

                                                         ^

exit status 1
invalid use of void expression

This report would have more information with
"Show verbose output during compilation"
option enabled in File -> Preferences.

Main.ino

    const int buzer = A0, on_off_pin = A1;

unsigned long current_millis;
const int time_interval = 20;
int currentMole = 0;

int points = 0, lives = 3, level = 0, odds_of_pop_up = 6;

mole moles[6] = {
  mole( 2, 8  ),
  mole( 3, 9  ),
  mole( 4, 10 ),
  mole( 5, 11 ),
  mole( 6, 12 ),
  mole( 7, 13 )
};
int number_of_mole = sizeof(moles) / sizeof(*moles);

    void timelyCall(unsigned long currentMillis, void (&f)(), int ms) {
  if (millis() - currentMillis > ms) {
    f();
    currentMillis = millis();
  }
}
void popAMole() {
  if (moles[currentMole].isReadyToPop()) {
    moles[currentMole].popUp(odds_of_pop_up);
  }
  if (currentMole + 1 <=  number_of_mole) {
    currentMole = 0;
  }
  else {
    currentMole++;
  }
}

mole.h

#ifndef mole_h
#define mole_h

#include "Arduino.h"

class mole
{
  public:
    mole(int p_out, int p_in);
    void resetMole();
    void popUp(int odds);
    boolean moleGetHit();
    boolean isReadyToPop();

    void setUpTimer(int up_timer);
    int  getUpTimer();
    void decrementUpTimer();

    void setRecoveryTimer(int recovery_timer);
    int  getRecoveryTimer();
    void decrementRecoveryTimer();
    void addRecoveryTimer(int add_time);

    boolean getSouldBeUp();

    int getOddsOfPopinh();
    void setOddsOfPopinh(int odds);

    int getPinIn();


  private:
    boolean _sould_be_up = false;
    int _up_timer = 0;
    int _recovery_timer = 0;
    int _pin_out;
    int _pin_in;
    int _odds_of_poping = 10; 

};
#endif

mole.cpp

boolean mole::isReadyToPop() {
  if (_recovery_timer == 0 && _sould_be_up = false) {
    return true;
  }
  else {
    return false;
  }
}

void mole::popUp(int odds) {
  if (rand() % odds + 1 == 1) {
    digitalWrite(_pin_out, HIGH)
    _sould_be_up = true;
  }
}

Answer 1

您正在调用未将地址传递给 timelyCall!

的函数

timelyCall(current_millis, popAMole(), time_interval);
timelyCall(current_millis, toggleBuzer(), 1000);

应该是：

timelyCall(current_millis, popAMole, time_interval);
timelyCall(current_millis, toggleBuzer, 1000);

请注意 popAMole 和 toggleBuzer 上没有括号。

无效使用 void 表达式，怎么办？

invalid use of void expression, what how?

c++

class

function

arduino

void