java - 返回方法值的类

Question

public class car
{
private int model;
private String make;
private double speed;
/**
 * *  My constructor
 */
public car()
{
    model = 2000;
    make= "Ferrari";
    speed= 50;

}

public int getYear()
{
    return model;
}

public String getMake()
{
    return make;
}

public double getSpeed()
{
    return speed;
}

public double accerlate ()
{
    double accerlate = speed++;
    return accerlate;
}

public void output(double accerlate)
{
    System.out.println("Year: " +model);
    System.out.println("Make: " + make);
    System.out.println("Speed: " + speed);
    System.out.println(accerlate);
}
}

连接到此类

public class RaceTrack
{
public static void main(String args[]){
   double speed = 50;
   car mycar= new car();
    mycar.getYear();
    mycar.getMake();
    mycar.getSpeed();
    mycar.output(speed);
    mycar.accerlate();
}
}

无论如何，我尝试获得输出，但加速方法无法按我想要的方式工作。应该是加+1，但它不起作用。为什么它不起作用？它只是打印原始速度。

Answer 1

双倍加速 = speed++; ==> 加速 =speed 然后速度 =speed+1

双倍加速 =++speed; ==> 加速 =speed+1

对于您的代码:你的输出将是:

Year: 2000
Make: Ferrari
Speed: 50.0
50.0

这是mycar.output(speed)的输出；其他方法还没有在你的main中输出

double speed = 50;
car mycar= new car();
 mycar.getYear(); //getYear return an int (model)=2000 so store it if you want
mycar.getMake(); //getMake return a string 
mycar.getSpeed();  //getSpeed return a double =50

如果你想看到加速的+1，你应该获取该方法的返回值然后打印它:

   Double ouputSpeed=mycar.accerlate();
   System.out.println(ouputSpeed.toString()); // print 51.0 

返回:

public double accerlate ()
{
    double accerlate = ++speed;
    return accerlate;
} 

==> 该方法返回一个 double 值 (accerlate= spped+1 =51)

==> 要在 main 中获取它，请声明一个 double : Double ouputSpeed=mycar.accerlate(); 然后打印它。