使用map.get(Object)时TreeMap返回null

Question

使用get方法获取值时, TreeMap将值打印为null,而它正常工作HashMap().请在下面找到示例代码并为此提供输入.

它适用于Hashmap,因为它使用equals()/hashcode()方法,而TreeMap是SortedMap,它不使用equals方法来比较两个对象.相反,它使用比较器/可比较来比较对象,但在使用get方法获取对象时,它将null作为响应.请在此提供一些清晰度.

    import java.util.Comparator;
    import java.util.HashMap;
    import java.util.Map;
    import java.util.Map.Entry;
    import java.util.TreeMap;

    class Employees implements Comparable<Employees>, Comparator<Employees> {

        public Employees(String name, int id) {
            super();
        this.name = name;
        this.id = id;
    }

    private String name;
    private int id;

    public String getName() {
        return name;
    }

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

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (!(obj instanceof Employees))
            return false;
        Employees other = (Employees) obj;
        if (id != other.id)
            return false;
        return true;
    }

    @Override
    public int hashCode() {
        return 1;
    }

    @Override
    public int compareTo(Employees o) {
        return 1;
    }

    @Override
    public int compare(Employees o1, Employees o2) {
        return 1;
    }
}

public class Employee {

    public static void main(String[] args) {
        // Map<Employees, Integer> m = new HashMap<>(); // equals will be used here.
        Map<Employees, Integer> m = new TreeMap<Employees, Integer>(); // no equals/hashcode used here as we use comparator/comparable to compare objects.
        Employees e1 = new Employees("abc", 11);
        Employees e2 = new Employees("abc", 12);
        System.out.println(m.put(e1, 1));
        System.out.println(m.put(e2, 2));
        **System.out.println(m.get(e2));**
    }
}

Answer 1

你的compareTo方法总是返回1,这意味着没有任何Employees对象等于任何其他Employees对象(甚至不是自己)compareTo.

仅在compareTo返回时0,两个比较实例被认为是相同的TreeMap.

顺便说一下,你hashCode总是回来的实施1也很糟糕.当它工作(for HashMap)时,它会导致所有条目存储在同一个桶中,从而破坏了它的性能HashMap.