java - Arrays.copyOf 在使类不可变方面扮演什么角色？

Question

Java 并发实践的摘录/片段-

@Immutable
class OneValueCache{
    private final BigInteger lastNumber;
    private final BigInteger[] lastFactors;

    public OneValueCache(BigInteger i, BigInteger[] factors) {
        this.lastNumber = i;
        this.lastFactors = Arrays.copyOf(factors,factors.length);     // 1
    }

    public BigInteger[] getFactors(BigInteger i){
        if(lastNumber == null || lastNumber.equals(i))
            return null;
        else
            return Arrays.copyOf(lastFactors,lastFactors.length);     // 2
    }
}

作者提到

OneValueCache wouldn't be immutable without the copyOf calls in the constructor and getter. Arrays.copyOf was added as a convenience in Java 6; clone would also work.

1) Arrays.copyOf 在使上述类成为 IMMUTABLE 方面扮演什么角色？ 如何？

上面的不可变类是这样使用的-

@ThreadSafe
public class VolatileCachedFactorizer implements Servlet{

    private volatile OneValueCache cache = new OneValueCache(null, null);

    public void service(ServletRequest req, ServletResponse resp){
        BigInteger i = extractFromRequest(req);
        BigInteger[] factors = cache.getFactors(i);

        if(factors == null){
         factors = factor(i);
         cache = new OneValueCache(i, factors); // cache being volatile, hence @ThreadSafe
        }

        encodeIntoResponse(resp, factors);
    }

}

Answer 1

如果您返回 this.lastFactors 而不是返回一个副本，调用者可以这样做(例如)

BigInteger[] lastFactors = cache.getFactors(...);
for (int i = 0; i < lastFactors.length; i++) {
    lastFactors[i] = null;
}

从而改变缓存的状态，这应该是不可变的。

构造函数的解释类似。如果构造函数没有复制，调用者可以做

factors = factor(i);
cache = new OneValueCache(i, factors); 
for (int i = 0; i < lastFactors.length; i++) {
    factors[i] = null;
}

因此再次改变缓存的状态。

经验法则:数组始终是可变的(空数组除外)。因此，如果一个不可变类的状态包含一个数组，那么调用者一定不能拥有对该数组的引用。