java - concurrentHashMap源码中的readValueUnderLock(e)存在的意义？

Answer 1

e.hash == hash && key.equals(e.key)

The previous sentence explains that when there is this key的，你看看put方法，当value为null in the table, an exception will occur:

public V put(K key, V value) {
        Segment<K,V> s;
        if (value == null)
            throw new NullPointerException();
        int hash = hash(key);
        int j = (hash >>> segmentShift) & segmentMask;
        if ((s = (Segment<K,V>)UNSAFE.getObject          // nonvolatile; recheck
             (segments, (j << SSHIFT) + SBASE)) == null) //  in ensureSegment
            s = ensureSegment(j);
        return s.put(key, hash, value, false);
    }

This is a higher version of mine. It may be different from yours, but the value cannot be empty.
So:

if (v != null)
    return v;
return readValueUnderLock(e); // recheck

If it is not empty, it can be returned directly. If it is empty, it means that other threads are operating it. So I added it.

public V get(Object key) {
        Segment<K,V> s; // manually integrate access methods to reduce overhead
        HashEntry<K,V>[] tab;
        int h = hash(key);
        long u = (((h >>> segmentShift) & segmentMask) << SSHIFT) + SBASE;
        if ((s = (Segment<K,V>)UNSAFE.getObjectVolatile(segments, u)) != null &&
            (tab = s.table) != null) {
            for (HashEntry<K,V> e = (HashEntry<K,V>) UNSAFE.getObjectVolatile
                     (tab, ((long)(((tab.length - 1) & h)) << TSHIFT) + TBASE);
                 e != null; e = e.next) {
                K k;
                if ((k = e.key) == key || (e.hash == h && key.equals(k)))
                    return e.value;
            }
        }
        return null;
    }

The current versionget方法把HashEntry<K,V> e弄成UNSAFE.getObjectVolatile()获取，像是volatile is out

Answer 2

The poster can upgrade the JDK, I use it 1.8找了一下，没发现这段代码在ConcurrentHashMap.

Answer 3

The object is null when it is initialized. This null is not assigned specifically in the program.