为什么当我输入一个整数时，我的 Java 双精度变量类型变成 0.0？

Question

我正在使用驱动程序 class 创建另一个 class 的对象。当我输入宠物重量或整数时，数字显示为 0.0。所有权重变量都声明为双精度变量，所以我不知道为什么要这样做。

  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Class

 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的事情是你只需要 class "Pet" 的一个实例。一只宠物可以容纳您需要的所有变量。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.");

当您只需要一个容器时，您编写它的方式实质上会创建 4 个不同的容器。您将权重分配给 petWeight 容器，但随后尝试从 petName 容器获取权重，这导致您检索到错误的变量。在这种情况下，只有一个名为 "pet" 的容器或实例，不应出现此问题。

Answer 2

如果 age 或 weight 没有预期的格式，我建议检查 getAge() 和 getWeight 的 return 类型。如果它与字段不匹配，它将在 returned 时被转换，至少在 int->double 的情况下（我相信 double->int 需要转换）。

但如前所述，添加 .0 是 double 的预期行为。如果你不想要它，你可以显式地转换为 int.

Answer 3

我看到的问题是您正在创建多个 Pet 对象并将宠物重量分配给 "petName" 对象，然后您试图将 "petWeight" 的重量添加到输出中目的。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.");         
    }
}

此外，我建议只使用一个对象 "pet" 并将每个值分配给那个对象，以及在 system.out

上只使用那个对象

Answer 4

问题出在第34行，看看你在做什么

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？您正在显示“petWeight”的宽度，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.

当然其他属性都是空的

希望有用。问候！

Answer 5

错误的是你用这个代码设置值

 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、检索中对 2 个语句使用相同的对象通过输入

petWeight.setWeight(lbs);

而不是

petName.setWeight(lbs);

会解决的

Answer 6

您在设置和获取您宠物的重量方面做错了。您需要使用同一个对象来设置和获取值。您正在 petName 对象中设置宠物重量并从 petWeight 对象获取。这就是为什么您会得到 0.0，这是 double 的默认值。解决你的问题使用相同的对象来设置和获取宠物的体重。

示例：

petWeight.setWeight(lbs); //setting the value in petWight object
petWeight.getWeight(); // it will returns the same value that you set.

Answer 7

感谢大家的帮助。犯了这么简单的错误，我觉得自己像个白痴。我还用 pet class.

的一个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.");         
        }
    }

为什么当我输入一个整数时，我的 Java 双精度变量类型变成 0.0？

Why is my Java double variable type coming out to 0.0 when I put in an integer?

java

accessor