当在JVM5和JVM6中运行相同的程序时,HashMap中的项目顺序不同

Question

我有一个应用程序,它显示行中的对象集合,一个对象=一行.对象存储在HashMap中.行的顺序不会影响应用程序的功能(这就是使用HashMap而不是可排序集合的原因).

但是我注意到,当使用两个不同版本的Java虚拟机运行时,相同的应用程序运行方式不同.应用程序使用JDK 5编译,可以使用Java 5或Java 6运行时运行,没有任何功能差异.

有问题的对象覆盖java.lang.Object#hashCode()并且显然需要遵循Java API中指定的契约.这可以通过以下事实得到证明:它们在应用程序的每次运行中始终以相同的顺序出现(在同一Java运行时中).

出于好奇,为什么Java运行时的选择会影响订单？

Answer 1

HashMap可以改变的实施细节.这个包私有方法最有可能(这是来自JDK 1.6.0_16):

/**
 * Applies a supplemental hash function to a given hashCode, which
 * defends against poor quality hash functions.  This is critical
 * because HashMap uses power-of-two length hash tables, that
 * otherwise encounter collisions for hashCodes that do not differ
 * in lower bits. Note: Null keys always map to hash 0, thus index 0.
 */
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);
}

作为参考,JDK 1.5.0_06中的模拟是:

/**
 * Returns a hash value for the specified object.  In addition to 
 * the object's own hashCode, this method applies a "supplemental
 * hash function," which defends against poor quality hash functions.
 * This is critical because HashMap uses power-of two length 
 * hash tables.<p>
 *
 * The shift distances in this function were chosen as the result
 * of an automated search over the entire four-dimensional search space.
 */
static int hash(Object x) {
    int h = x.hashCode();

    h += ~(h << 9);
    h ^=  (h >>> 14);
    h +=  (h << 4);
    h ^=  (h >>> 10);
    return h;
}

Answer 2

可能是因为a Map未定义为具有任何特定的迭代顺序; 元素返回的顺序可能是其内部实现的工件,并且不需要保持一致.

如果在Java 5和6之间更新实现(特别是出于性能原因),Sun没有任何好处或义务确保迭代顺序在两者之间保持一致.

编辑:我刚刚在Java 6早期版本中发现了一个有趣的片段(不幸的是我不确定它的确切版本,但显然是2006年6月的HashMap 1.68):

 /**
  * Whether to prefer the old supplemental hash function, for
  * compatibility with broken applications that rely on the
  * internal hashing order.
  *
  * Set to true only by hotspot when invoked via
  * -XX:+UseNewHashFunction or -XX:+AggressiveOpts
  */
 private static final boolean useNewHash;
 static { useNewHash = false; }

 private static int oldHash(int h) {
     h += ~(h << 9);
     h ^= (h >>> 14);
     h += (h << 4);
     h ^= (h >>> 10);
     return h;
 }

 private static int newHash(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);
 }

所以看起来尽管我上面的断言,Sun实际上确实考虑了迭代顺序的一致性 - 在稍后的某个时候,这个代码可能会被丢弃,而新的命令也是最终的.