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ConcurrentHashMap.initTable(), why check table is null twice?

I'm learning the java source code, when i reading the ConcurrentHashMap source code, i'm confused with the initTable() method, why check (tab = table) == null || tab.length == 0 twice, first in the while(), then in the if(). I can't imagine in what situation need the second check.

I think maybe it's because the JVM reorder the code, put sizeCtl = sc; in front of Node<K,V>[] nt = (Node<K,V>[])new Node<?,?>[n];. It's just my guess, I don't know is it right.

Can some one explain it, thanks a lot.

 private final Node<K,V>[] initTable() {
        Node<K,V>[] tab; int sc;
        while ((tab = table) == null || tab.length == 0) {
            if ((sc = sizeCtl) < 0)
                Thread.yield(); // lost initialization race; just spin
            else if (U.compareAndSwapInt(this, SIZECTL, sc, -1)) {
                try {
                    if ((tab = table) == null || tab.length == 0) {
                        int n = (sc > 0) ? sc : DEFAULT_CAPACITY;
                        @SuppressWarnings("unchecked")
                        Node<K,V>[] nt = (Node<K,V>[])new Node<?,?>[n];
                        table = tab = nt;
                        sc = n - (n >>> 2);
                    }
                } finally {
                    sizeCtl = sc;
                }
                break;
            }
        }
        return tab;
    }
over 4 years ago · Santiago Trujillo
1 answers
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0

Multiple threads may compete to do this (see "initialization race" comment).

To paraphrase the code:

while(uninitialized) {
    acquire_lock(); //compareAndSwapInt...
    if(uninitialized) {
        do_init();
    }
}

The outer check is a cheap "unlocked" test. The inner one is in case someone else already succeeded between while and compareAndSwapInt.

over 4 years ago · Santiago Trujillo Report
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