我的问题更多是适用于这个特定实例的概念/学习问题。我正在做作业,我有一个名为 sensor 的类和一个名为 digitalSensor 的派生类。传感器的数据成员之一是“运行”。而且,当我实现 digitalSensor 的打印功能时,我需要根据数字传感器是否“正常工作”打印出一行。
本质上,我需要在 digitalSensor 中创建一个 if 语句来检查“运行”的值。但是 Xcode 告诉我“功能是传感器的私有成员”。由于 digitalSensor 是从传感器派生的,它不应该也有“功能”成员变量吗?如何在创建 digitalSensor 打印功能时检查它?
这是我的 sensor.h 文件:
#ifndef __Program_6__sensor__
#define __Program_6__sensor__
#include <iostream>
class sensor {
char* SensorName;
float energyDraw;
int functioning;
int onoff;
public:
sensor(char*n, float pc);
virtual void print();
void setOK(int K);
int getOK();
void setOnOff(int n);
int getOnOff();
};
//---------
class digitalSensor : public sensor {
int reading;
public:
digitalSensor(char*n, float pc);
virtual void print();
void setCurrentReading(int r);
int getCurrentReading();
};
class analogSensor : public sensor {
int Reading;
int minRead;
int maxRead;
public:
analogSensor(char *n, float pc, int mm, int mx);
virtual void print();
void setCurrentReading(int r);
int getCurrentReading();
};
#endif /* defined(__Program_6__sensor__) */
这是我的 sensor.cpp 文件,我在打印功能上的工作在底部:
#include "sensor.h"
#include "definitions.h"
using namespace std;
//--------SENSOR CLASS------------//
sensor::sensor(char *n, float pc) {
SensorName = (char*)malloc(strlen(n)+1);
energyDraw = pc;
functioning = WORKING;
onoff = OFF;
}
void sensor::print() {
cout << " Sensor: " << SensorName;
cout << " Power Consumption: " << energyDraw;
if (functioning == WORKING) {
cout << "\nSensor is functioning correctly\n";
if (onoff == ON) {
cout << "Sensor is On";
}
if (onoff == OFF) {
cout << "Sensor is Off";
}
}
if (functioning == NOTWORKING) {
cout << "Sensor is not functioning correctly";
}
}
void sensor::setOK(int k) {
functioning = k;
}
int sensor::getOK() {
return functioning;
}
void sensor::setOnOff(int n) {
onoff = n;
}
int sensor::getOnOff() {
return onoff;
}
//---------------------------------//
//*********DIGITAL SENSOR**********//
digitalSensor::digitalSensor(char *n, float pc) : sensor(n, pc){
reading = OFF;
}
void digitalSensor::print() {
sensor::print();
if (functioning == WORKING && onoff == ON) {
cout << "Current sensor reading is: " << reading;
}
if (digitalSensor.functioning == WORKING && digitalSensor.onoff == OFF) {
cout << "Current reading not available";
}
}
错误是最后的“void digitalSensor::print()”之后的两行。
感谢您提供的任何帮助!一旦我学会了这些东西,我一定有一天也会回报和回答新手的问题!