标记“***”上出现语法错误，此标记后面应为AnnotationName[已关闭]

Question

我本打算在我的代码下设置狗的高度，但它用上面的错误标记了它。是否有任何修复方法，是否缺少任何{}im？

就像代码差不多完成了一样，我检查了所有代码，它工作了，并尝试添加狗的高度，但它抛出了语法错误

public class Dog
{
private int age;
private String name;
private String Color;
private int Height;
public Dog(int age,String name,String Color, int Height)
{
   this.age= age;
   this.name= name;
   this.Color = Color;
   this. Altura = Altura;
}
public String setDogName(String Name)
{
   this.name = name;
   return name;
}
public String getDogName()
{
   return name;
}
public String setDogColor(String Color)
{
   this.Color = Color;
   return Color;

}
{
   public int setDogHeight()
}
public int setDogAge(int age)
{
   this.age = age;
   return age;
}

public int getDogAge()
{
   return age;
}

public int computeDogAge()
{
   this.age=age*7;
   return age;
}
public String toString()
{
   String dogsname= "Dog's name:";
   String dogsage= "Dog's age:";
   String dogcolor= "Dog's Color:";
   return dogsname + name +"\t"+dogsage+age+"\t"+dogcolor+Color;
}

public static void main (String[] args)
{
   Dog dog1 = new Dog (1,"sasha","red");
   Dog dog2 = new Dog (7,"karmine","green");
   Dog dog3 = new Dog (6,"dakota","blue");

   dog1.setDogAge(9);
   System.out.println((dog1));

   dog2.setDogName("shazam");
   System.out.println(dog2);

   System.out.println(dog3);

   System.out.println();
   System.out.println("Dog's age in human years:");
   System.out.println(dog1.computeDogAge());
   System.out.println(dog2.computeDogAge());
   System.out.println(dog3.computeDogAge());
   
   if(dog1.age >dog2.age); {
       System.out.println( "sasha is older than karmine");

   }

   
}
}

错误出现在第28到31行，代码如下

}
{
    public int setDogHeight()
}

Answer 1

有一些较小的类型错误：

• 你忘了把altura改成高度
• 方法声明的括号后面有一个冒号
• 在main（）中创建的Dog不遵循您的构造函数
• setdoghight（）的花括号放在方法声明的周围
• setDogName（）中的属性name已大写

以下是更正后的代码：

package com.company;

public class Dog {
    private int age;
    private String name;
    private String color;
    private int height;

    public Dog(int age, String name, String color, int height) {
        this.age = age;
        this.name = name;
        this.color = color;
        this.height = height;
    }

    public void setDogName(String name) {
        this.name = name;
    }

    public String getDogName() {
        return name;
    }

    public void setDogColor(String color) {
        this.color = color;
    }

    public void setDogHeight(int height) {
        this.height = height;
    }

    public void setDogAge(int age) {
        this.age = age;
    }

    public int getDogAge() {
        return age;
    }

    public int computeDogAge() {
        this.age = age * 7;
        return age;
    }

    public String toString() {
        String dogName = "Dog's name: ";
        String dogAge = "Dog's age: ";
        String dogColor = "Dog's Color: ";
        return dogName + name + "\t" + dogAge + age + "\t" + dogColor + color;
    }

    public static void main(String[] args) {
        Dog dog1 = new Dog(1, "sasha", "red", 1);
        Dog dog2 = new Dog(7, "karmine", "green", 1);
        Dog dog3 = new Dog(6, "dakota", "blue", 1);

        dog1.setDogAge(9);
        System.out.println((dog1));

        dog2.setDogName("shazam");
        System.out.println(dog2);

        System.out.println(dog3);

        System.out.println();
        System.out.println("Dog's age in human years:");
        System.out.println(dog1.computeDogAge());
        System.out.println(dog2.computeDogAge());
        System.out.println(dog3.computeDogAge());

        if (dog1.age > dog2.age) {
            System.out.println(dog1.name + " is older than " + dog2.name);
        }
    }
}

避免这些问题的一些建议：

>

遵守代码约定，例如使用小写字符作为变量和方法名。

不要存储任何信息两次。在main（）中的if-语句中，您打印出“萨沙比卡敏年长”。但是当你创建它们的时候，它们的名字已经被定义了。现在，即使您更改了名称，程序也不会打印“shazam”而不是“karmine”。改为访问属性（请参见代码）。

不要更改getter中的代码，比如computeDogAge（）。如果你第二次调用这个方法，它会再次乘以年龄。相反，您可以直接返回age*7（参见代码）。

如果你格式化你的代码，你就不会再有像这样的编译错误了。使用像IntelliJ这样的IDE，使用StrgAltL格式化代码。