Question

我正在使用驱动程序类来创建另一个类的对象。当我输入宠物重量或整数时，数字变为0.0。所有的权重变量都被声明为double，所以我不知道为什么会这样做。

  import java.util.Scanner;

    public class PetAssignment {

        public static void main(String[] args) 
        {
            String nameAndType;
            int yrs;
            double lbs;

            //Scanner object for the keyboard input
            Scanner answers = new Scanner(System.in);

            //Pet objects used for calling accessor methods
            Pet petName = new Pet();
            Pet petType = new Pet();
            Pet petAge = new Pet();
            Pet petWeight = new Pet();

            //A bunch of other code and pet attributes

            //Input for the weight of pet
            System.out.print("How many pounds does your pet weight? ");
            lbs = answers.nextDouble();
            petName.setWeight(lbs);

            //Print out of the user's answers
            System.out.println(""); 
            System.out.println("You have a "+ petType.getType() + ". That is named " 
                + petName.getName()+ " and is " 
                + petAge.getAge() + " years old and weighs " 
                + petWeight.getWeight() + " lbs.");         
        }
    }

这是我的宠物类

 public class Pet 
    {
      private String name;
      private String type;
      private int age;
      private double weight;    
      /*
       * a bunch of other code
       */
      public void setWeight(double petWeight)
      {
          weight = petWeight;
      }
      /*
       * a bunch of other code
       */
      public double getWeight()
      {
          return weight;
      }
    }

Answer 1

这个问题的第一个重要问题是你只需要一个“Pet”类的实例。一只宠物可以容纳你需要的所有变量。例如：

Pet pet = new Pet();
System.out.print("How many pounds does your pet weight? ");
lbs = answers.nextDouble();
petName.setWeight(lbs);

 System.out.println(""); 
        System.out.println("You have a "+ pet.getType() + ". That   is named " 
            + pet.getName()+ " and is " 
            + pet.getAge() + " years old and weighs " 
            + pet.getWeight() + " lbs.");

你写它的方式基本上只需要一个就可以创建4个不同的容器。您已将权重分配给petWeight容器，但随后尝试从petName容器中获取权重，这导致您检索错误的变量。在这种情况下，只有一个容器或实例称为“pet”，不应发生此问题。

Answer 2

错误的是您使用此代码设置值

 petName.setWeight(lbs);

并使用它来检索值

 Pet petWeight = new Pet();

它们是 2个不同的对象您必须在设置和检索中对2个语句使用相同的对象通过推荐

petWeight.setWeight(lbs);

而不是

petName.setWeight(lbs);

将解决它

Answer 3

如果age或weight没有预期的格式，我建议您检查getAge()和getWeight的返回类型。如果它与字段不匹配，则会在返回时进行转换，至少在int->double情况下（我相信double->int需要转换）。

但如前所述，添加.0是double的预期行为。如果您不想要，可以明确转换为int。

Answer 4

我看到的问题是你正在创建多个Pet对象并将宠物重量分配给＆＃34; petName＆＃34;对象然后你试图在输出中添加＆＃34; petWeight＆＃34;的重量。宾语。请尝试以下操作：

        //Input for the weight of pet
        System.out.print("How many pounds does your pet weight? ");
        lbs = answers.nextDouble();
        petWeight.setWeight(lbs);

        //Print out of the user's answers
        System.out.println(""); 
        System.out.println("You have a "+ petType.getType() + ". That is named " 
            + petName.getName()+ " and is " 
            + petAge.getAge() + " years old and weighs " 
            + petWeight.getWeight() + " lbs.");         
    }
}

另外，我建议只使用一个物品＆＃34; pet＆＃34;并将每个值分配给该值，并在system.out

上仅使用该值

Answer 5

问题出在第34行，检查你在这做什么

petName.setWeight(lbs);

但是当你显示输出

时

System.out.println("You have a "+ petType.getType() + ". That is named " 
            + petName.getName()+ " and is " 
            + petAge.getAge() + " years old and weighs " 
            + petWeight.getWeight() + " lbs.");

你看到了吗？你正在显示＆＃34; petWeight的wieight，但扫描仪正在设置petName对象，请调试并检查它。

System.out.print("How many pounds does your pet weight? ");
lbs = answers.nextDouble();
petWeight.setWeight(lbs);

结果是

 You have a null. That is named null and is 0 years old and weighs 1.5 lbs.

其他属性为空。

我希望这很有用。此致！

Answer 6

感谢大家的帮助。有这么简单的错误，我觉得自己像个白痴。我还用一个pet class实例简化了我的代码。

import java.util.Scanner;

    public class PetAssignment {

        public static void main(String[] args) 
        {
            String nameAndType;
            int yrs;
            double lbs;

            //Scanner object for the keyboard input
            Scanner answers = new Scanner(System.in);

            //Pet objects used for calling accessor methods
            Pet pet = new Pet();

            //A bunch of other code and pet attributes

            //Input for the weight of pet
            System.out.print("How many pounds does your pet weight? ");
            lbs = answers.nextDouble();
            pet.setWeight(lbs);

            //Print out of the user's answers
            System.out.println(""); 
            System.out.println("You have a "+ pet.getType() + ". That is named " 
                + pet.getName()+ " and is " 
                + pet.getAge() + " years old and weighs " 
                + pet.getWeight() + " lbs.");         
        }
    }

Answer 7

你在设置和获得你的宠物重量方面做错了。您需要使用相同的对象来设置和获取值。您正在petWeight对象中设置宠物重量并从0.0对象获取。这就是为什么您获得double这是petWeight.setWeight(lbs); //setting the value in petWight object petWeight.getWeight(); // it will returns the same value that you set.的默认值的原因。解决你的问题使用相同的物体来设定并获得宠物的体重。

示例：

{{1}}

当我输入一个整数时，为什么我的Java双变量类型会出现0.0？

7 个答案: