Question

***我遇到了正确创建add（）subtract（）和divide（）方法的问题，也许我正在过度思考它。我如何正确返回所需的对象。我的IDE说“无法解析符号温度”。我正确地称这个'这个'了吗？

***我知道add和divide方法是不完整的，如果我可以在完成subtract（）方法时获得帮助，那么一切都应该落实到位。

    import javax.naming.StringRefAddr;

/**
 * Created by Makoto on 2/4/2016.
 */

public class Temperature {
    public double temp;
    public char type;

    Temperature() {
        temp = 0.0;
        type = 'o';

    }

    Temperature( double temperature, char tempType){
            temperature = temp;
            tempType = type;
    }

    public Temperature toFahrenheit() {
        type.equalsIgnoreCase();
        switch (type) {
            case 'C':

                break;

            case 'K':

                break;

            default:
                System.out.println("Can not convert.");
                break;
        }

        return this.Temperature;
    }

    public Temperature toCelsius() {
        type.equalsIgnoreCase();
        switch (type) {
            case 'F':

                break;

            case 'K':

                break;

            default:
                System.out.println("Can not convert.");
                break;
        }

        return this.Temperature;
    }

    public Temperature toKelvin() {
        type.equalsIgnoreCase();
        switch (type) {
            case 'C':

                break;

            case 'F':

                break;

            default:
                System.out.println("Can not convert.");
                break;
        }

        return this.Temperature;
    }

    public Temperature add(Temperature x){

        this.Temperature + x;
        return this.Temperature;
    }

    public Temperature subtract(Temperature x){

        this.Temperature = this.temp - x.temp;
        return this.Temperature;
    }
    public Temperature divide(int x){

        this.Temperature = /x;
        return this.Temperature;
    }

}

Answer 1

您只需要退货。

  import javax.naming.StringRefAddr;

/**
 * Created by Makoto on 2/4/2016.
 */

public class Temperature {
    public double temp;
    public char type;

    Temperature() {
        temp = 0.0;
        type = 'o';

    }

    Temperature( double temperature, char tempType){
            temperature = temp;
            tempType = type;
    }

    public Temperature toFahrenheit() {
        type.equalsIgnoreCase();
        switch (type) {
            case 'C':

                break;

            case 'K':

                break;

            default:
                System.out.println("Can not convert.");
                break;
        }

        return this;
    }

    public Temperature toCelsius() {
        type.equalsIgnoreCase();
        switch (type) {
            case 'F':

                break;

            case 'K':

                break;

            default:
                System.out.println("Can not convert.");
                break;
        }

        return this;
    }

    public Temperature toKelvin() {
        type.equalsIgnoreCase();
        switch (type) {
            case 'C':

                break;

            case 'F':

                break;

            default:
                System.out.println("Can not convert.");
                break;
        }

        return this;
    }

    public Temperature add(Temperature x){

        this.temp + x.temp;
        return this;
    }

    public Temperature subtract(Temperature x){

        this.temp = this.temp - x.temp;
        return this;
    }
    public Temperature divide(int x){

        this.temp = /x;
        return this;
    }

}

Answer 2

this.Temperature = this.temp - x.temp;不是Java中的有效语句。您试图将Temperature（它本身不是对象）设置为值。您在此处尝试执行的操作是对temp变量执行操作，然后返回this。通过返回this，调用对象现在将具有带有更新字段的Temperature对象。

例如：

public Temperature subtract(Temperature x){ temp = temp - x.temp; return this; }

上面的代码将更新此温度对象的临时，然后返回自身。然后，调用对象可以通过调用returnedTemperatureObject.temp来访问它。

无法解析温度驱动程序的Java类中的方法

2 个答案: