是不是抽象的，不重寫抽象方法行爲

Question

我有一個小模擬器，模擬狐狸和兔子... 我想讓狐狸和兔子類來實現演員類，但我得到這個錯誤消息，我不知道什麼是錯..

錯誤消息：兔子是不是抽象的，在演員不重寫抽象方法步驟（java.until.List）

兔類

import java.util.List; 
import java.util.Random; 


public class Rabbit extends Animal implements Actor 
{ 
    // Characteristics shared by all rabbits (static fields). 

    // The age at which a rabbit can start to breed. 
    private static final int BREEDING_AGE = 5; 
    // The age to which a rabbit can live. 
    private static final int MAX_AGE = 40; 
    // The likelihood of a rabbit breeding. 
    private static final double BREEDING_PROBABILITY = 0.15; 
    // The maximum number of births. 
    private static final int MAX_LITTER_SIZE = 4; 

    /** 
    * Create a new rabbit. A rabbit may be created with age 
    * zero (a new born) or with a random age. 
    * 
    * @param randomAge If true, the rabbit will have a random age. 
    * @param field The field currently occupied. 
    * @param location The location within the field. 
    */ 
    public Rabbit(boolean randomAge, Field field, Location location) 
    { 
     super(field, location); 
     if(randomAge) { 
      setAge(rand.nextInt(MAX_AGE)); 
     } 
    } 

    /** 
    * This is what the rabbit does most of the time - it runs 
    * around. Sometimes it will breed or die of old age. 
    * @param newRabbits A list to add newly born rabbits to. 
    */ 
    public void act(List<Actor> newActors) 
    { 
     incrementAge(); 
     if(isActive()) { 
      giveBirth(newRabbits);    
      // Try to move into a free location. 
      Location newLocation = getField().freeAdjacentLocation(getLocation()); 
      if(newLocation != null) { 
       setLocation(newLocation); 
      } 
      else { 
       // Overcrowding. 
       setDead(); 
      } 
     } 
    } 


    public Animal getNewAnimal(boolean status, Field field, Location loc) 
    { 
     return new Rabbit(status, field, loc); 
    } 

    /** 
    * Return the maximal age of the rabbit. 
    * @return The maximal age of the rabbit. 
    */ 
    protected int getMaxAge() 
    { 
     return MAX_AGE; 
    } 

    /** 
    * Return the breeding age of the rabbit. 
    * @return The breeding age of the rabbit. 
    */ 
    protected int getBreedingAge() 
    { 
     return BREEDING_AGE; 
    } 

    /** 
    * Return the breeding probability of the rabbit. 
    * @return The breeding probability of the rabbit. 
    */ 

    protected double getBreedingProbability() 
    { 
     return BREEDING_PROBABILITY; 
    } 

    /** 
    * Return the maximal litter size of the rabbit. 
    * @return The maximal litter size of the rabbit. 
    */ 

    protected int getMaxLitterSize() 
    { 
     return MAX_LITTER_SIZE; 
    } 

} 

演員類

import java.util.List; 
/** 
* Write a description of interface Actor here. 
* 
* @author (your name) 
* @version (a version number or a date) 
*/ 
public interface Actor 
{ 
    /** 
    * performs actor's regular behaviour 
    */ 
    void act(List<Actor> newActors); 
    /** 
    * is the actor still active 
    */ 
    boolean isActive(); 
} 

Answer 1

通過實施Actor，您承諾實施void act(List<Actor> xxx)方法，但您沒有。您確實實施了方法void act(List<Animal> xx)，但這是一種不同的方法（具有不同的簽名）。

簡單地說，你沒有正確重寫act方法：

public void act(List<Animal> newRabbits) != void act(List<Actor> newActors); 

簽名必須一致。由於act是Actor抽象（即接口），則必須將其覆蓋在實現Actor類或聲明它abstract（不能夠實例化。

Answer 2

接口是一個合同，你將在類提供某些功能實現這些接口您MUST實現從演員接口的行爲（）方法，如果你想創建一個具體類

你可以在這裏閱讀更多：。 http://docs.oracle.com/javase/tutorial/java/concepts/interface.html

Answer 3

@J ack是對的。

我想解釋爲什麼這是不允許的。 （如果它是一個數組將被允許。）

考慮動物類

public class Animal{ 
     String name; 

    //getters and setters 
} 

public class Dog extends Animal{ 

public void woof(){ } 
} 

public class Cat extends Animal{ 
    public void meow(){} 
} 

PS：以下是不允許的，但顯示只是爲了解釋爲什麼它不是允許

List<Animal> lst = new ArrayList<Animal>(); 

lst.add(new Dog()); 
lst.add(new Cat()); 

makeNoise(lst); //**** IS NOT ALLOWED BUT SIMULATES YOUR SITUATION. *****// 

你有一個方法在一些其他類是

public makeNoise(List<Dog> dogNoises){ 

//iterate over the list and do follwoing for each Dog 
//Remeber we put a cat in Dog list? Cat doesnt woof!!! it will break or crash or something else you name it. 

for(Dog doggie: dogNoises) 
    doggie.woof(); 
} 

通用集合確保我們不碰到是起訴如上。