問題描述

我分析Java中的HashMap源代碼，得到一個關于put方法的問題.

I analyse the HashMap source code in Java and get a question about the put method.

下面是JDK1.6中的put方法:

Below is the put method in JDK1.6:

public V put(K key, V value) {
    if (key == null)
        return putForNullKey(value);
    int hash = hash(key.hashCode());
    int i = indexFor(hash, table.length);
    for (Entry<K,V> e = table[i]; e != null; e = e.next) {
        Object k;
        if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
            V oldValue = e.value;
            e.value = value;
            e.recordAccess(this);
            return oldValue;
        }
    }

    modCount++;
    addEntry(hash, key, value, i);
    return null;
}

我對 if (e.hash == hash && ((k = e.key) == key || key.equals(k))) 感到困惑

為什么會這樣?

因為在Java超類Object中，有一個contract of hashCode and equals:

Because in Java super class Object, there is a contract of hashCode and equals:

如果兩個對象根據equals(Object)方法相等，那么對兩個對象中的每一個調用hashCode方法必須產生相同的整數結果.

If two objects are equal according to the equals(Object) method, then calling the hashCode method on each of the two objects must produce the same integer result.

所以 key.equals(k) 意味著 key.hashCode() == k.hashCode().

hash()如下:

 static int hash(int h) {
    // This function ensures that hashCodes that differ only by
    // constant multiples at each bit position have a bounded
    // number of collisions (approximately 8 at default load factor).
    h ^= (h >>> 20) ^ (h >>> 12);
    return h ^ (h >>> 7) ^ (h >>> 4);
}

因此 key.hashCode() == k.hashCode() 意味著 e.hash == hash.

那么為什么不是if ((k = e.key) == key || key.equals(k))這樣的條件?

推薦答案

這只是一個優化:比較兩個整數比調用 equals() 更快.

It's just an optimization: comparing two integers is faster than calling equals().

如果兩個hashcode不同，那么根據equals和hashCode的約定，map知道現有的key不等于給定的key，并且可以更快地進入下一個.

If the two hashcodes differ, then, based on the contract of equals and hashCode, the map knows that the existing key isn't equal to the given key, and can go faster to the next one.

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