理解HashMap中equals和hashCode的工作原理

HashMap使用hashCode()，==并equals()用于条目查找。给定键的查找序列k如下：使用k.hashCode()来确定条目存储其斗，如果有的话如果找到，对于k1该桶中的每个条目的密钥，如果k == k1 || k.equals(k1)，则返回k1的条目任何其他结果，没有相应的条目为了演示一个例子，假设我们想要创建一个HashMapwhere键，如果它们具有相同的整数值（由AmbiguousIntegerclass 表示），则它们在逻辑上是等价的。然后我们构造一个HashMap，放入一个条目，然后尝试覆盖其值并按键检索值。class AmbiguousInteger {    private final int value;    AmbiguousInteger(int value) {        this.value = value;    }}HashMap<AmbiguousInteger, Integer> map = new HashMap<>();// logically equivalent keysAmbiguousInteger key1 = new AmbiguousInteger(1),                 key2 = new AmbiguousInteger(1),                 key3 = new AmbiguousInteger(1);map.put(key1, 1); // put in value for entry '1'map.put(key2, 2); // attempt to override value for entry '1'System.out.println(map.get(key1));System.out.println(map.get(key2));System.out.println(map.get(key3));Expected: 2, 2, 2不要覆盖hashCode()和equals()：在默认情况下的Java生成不同的hashCode()不同对象的值，因此HashMap使用这些值映射key1和key2成不同的桶。key3没有相应的桶，所以它没有价值。class AmbiguousInteger {    private final int value;    AmbiguousInteger(int value) {        this.value = value;    }}map.put(key1, 1); // map to bucket 1, set as entry 1[1]map.put(key2, 2); // map to bucket 2, set as entry 2[1]map.get(key1); // map to bucket 1, get as entry 1[1]map.get(key2); // map to bucket 2, get as entry 2[1]map.get(key3); // map to no bucketExpected: 2, 2, 2Output:   1, 2, nullhashCode()仅覆盖： HashMap映射key1并key2进入同一个存储桶，但由于两者key1 == key2和key1.equals(key2)检查失败，它们保持不同的条目，因为默认情况下equals()使用==check，它们引用不同的实例。key3失败都==和equals()检查对key1和key2，因此具有没有对应的值。class AmbiguousInteger {    private final int value;    AmbiguousInteger(int value) {        this.value = value;    }    @Override    public int hashCode() {        return value;    }}map.put(key1, 1); // map to bucket 1, set as entry 1[1]map.put(key2, 2); // map to bucket 1, set as entry 1[2]map.get(key1); // map to bucket 1, get as entry 1[1]map.get(key2); // map to bucket 1, get as entry 1[2]map.get(key3); // map to bucket 1, no corresponding entryExpected: 2, 2, 2Output:   1, 2, nullequals()仅覆盖： HashMap由于默认值不同，将所有密钥映射到不同的存储桶hashCode()。==或者equals()检查在这里是无关紧要的，因为它HashMap永远不会达到需要使用它们的程度。class AmbiguousInteger {    private final int value;    AmbiguousInteger(int value) {        this.value = value;    }    @Override    public boolean equals(Object obj) {        return obj instanceof AmbiguousInteger && value == ((AmbiguousInteger) obj).value;    }}map.put(key1, 1); // map to bucket 1, set as entry 1[1]map.put(key2, 2); // map to bucket 2, set as entry 2[1]map.get(key1); // map to bucket 1, get as entry 1[1]map.get(key2); // map to bucket 2, get as entry 2[1]map.get(key3); // map to no bucketExpected: 2, 2, 2Actual:   1, 2, null覆盖both hashCode()和equals()：HashMapmaps key1，key2并覆盖key3到同一个存储桶中。==在比较不同的实例时检查失败，但是equals()检查通过，因为它们都具有相同的值，并且被我们的逻辑视为“逻辑上等效”。class AmbiguousInteger {    private final int value;    AmbiguousInteger(int value) {        this.value = value;    }    @Override    public int hashCode() {        return value;    }    @Override    public boolean equals(Object obj) {        return obj instanceof AmbiguousInteger && value == ((AmbiguousInteger) obj).value;    }}map.put(key1, 1); // map to bucket 1, set as entry 1[1]map.put(key2, 2); // map to bucket 1, set as entry 1[1], override valuemap.get(key1); // map to bucket 1, get as entry 1[1]map.get(key2); // map to bucket 1, get as entry 1[1]map.get(key3); // map to bucket 1, get as entry 1[1]Expected: 2, 2, 2Actual:   2, 2, 2如果hashCode()随机怎么办？：HashMap将为每个操作分配一个不同的存储桶，因此您永远不会找到您之前输入的相同条目。class AmbiguousInteger {    private static int staticInt;    private final int value;    AmbiguousInteger(int value) {        this.value = value;    }    @Override    public int hashCode() {        return ++staticInt; // every subsequent call gets different value    }    @Override    public boolean equals(Object obj) {        return obj instanceof AmbiguousInteger && value == ((AmbiguousInteger) obj).value;    }}map.put(key1, 1); // map to bucket 1, set as entry 1[1]map.put(key2, 2); // map to bucket 2, set as entry 2[1]map.get(key1); // map to no bucket, no corresponding valuemap.get(key2); // map to no bucket, no corresponding valuemap.get(key3); // map to no bucket, no corresponding valueExpected: 2, 2, 2Actual:   null, null, null如果hashCode()总是一样的话怎么办？：HashMap将所有键映射到一个大桶中。在这种情况下，您的代码在功能上是正确的，但使用HashMap实际上是多余的，因为任何检索都需要在O（N）时间内迭代该单个存储桶中的所有条目（或Java 8的O（logN）），等效使用a List。class AmbiguousInteger {    private final int value;    AmbiguousInteger(int value) {        this.value = value;    }    @Override    public int hashCode() {        return 0;    }    @Override    public boolean equals(Object obj) {        return obj instanceof AmbiguousInteger && value == ((AmbiguousInteger) obj).value;    }}map.put(key1, 1); // map to bucket 1, set as entry 1[1]map.put(key2, 2); // map to bucket 1, set as entry 1[1]map.get(key1); // map to bucket 1, get as entry 1[1]map.get(key2); // map to bucket 1, get as entry 1[1]map.get(key3); // map to bucket 1, get as entry 1[1]Expected: 2, 2, 2Actual:   2, 2, 2如果equals总是假的怎么办？：==当我们将同一个实例与自身进行比较时检查通过，但是否则失败，equals检查总是失败key1，key2并且key3被认为是“逻辑上不同”，并且映射到不同的条目，尽管它们仍然在同一桶中hashCode()。class AmbiguousInteger {    private final int value;    AmbiguousInteger(int value) {        this.value = value;    }    @Override    public int hashCode() {        return 0;    }    @Override    public boolean equals(Object obj) {        return false;    }}map.put(key1, 1); // map to bucket 1, set as entry 1[1]map.put(key2, 2); // map to bucket 1, set as entry 1[2]map.get(key1); // map to bucket 1, get as entry 1[1]map.get(key2); // map to bucket 1, get as entry 1[2]map.get(key3); // map to bucket 1, no corresponding entryExpected: 2, 2, 2Actual:   1, 2, null好的，如果equals现在总是如此？：你基本上是说所有对象都被认为与另一个对象“在逻辑上等同”，所以它们都映射到同一个桶（由于相同hashCode()），相同的条目。class AmbiguousInteger {    private final int value;    AmbiguousInteger(int value) {        this.value = value;    }    @Override    public int hashCode() {        return 0;    }    @Override    public boolean equals(Object obj) {        return true;    }}map.put(key1, 1); // map to bucket 1, set as entry 1[1]map.put(key2, 2); // map to bucket 1, set as entry 1[1], override valuemap.put(new AmbiguousInteger(100), 100); // map to bucket 1, set as entry1[1], override valuemap.get(key1); // map to bucket 1, get as entry 1[1]map.get(key2); // map to bucket 1, get as entry 1[1]map.get(key3); // map to bucket 1, get as entry 1[1]Expected: 2, 2, 2Actual:   100, 100, 100

理解HashMap中equals和hashCode的工作原理

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