java - 您能解释一下计算哈希码的所谓 Java 标准方法吗？

Question

我最近必须重写 Java 中的 equals 和 hashCode 方法。因此，我寻找一种快速有效的方法来计算哈希码。

Java 开发人员似乎同意以下方法:

    int hash = 23;
    hash = hash * 37 + paramOne;
    hash = hash * 37 + paramTwo;
    // And so on...

这可能是简单的算术，但我不太明白。有哪些保证？什么是角落案例？是否有更好(相当简单)的方法来做到这一点？

谢谢!

Answer 1

用 Joshua Bloch 的话(解释 String 类中 hashCode() method 的默认实现，即:s[0]*31 ^(n-1) + s[1]*31^(n-2) + ... + s[n-1]):

The value 31 was chosen because it is an odd prime. If it were even and the multiplication overflowed, information would be lost, as multiplication by 2 is equivalent to shifting. The advantage of using a prime is less clear, but it is traditional. A nice property of 31 is that the multiplication can be replaced by a shift and a subtraction for better performance: 31 * i == (i << 5) - i. Modern VMs do this sort of optimization automatically.

如需进一步阅读，请参阅 this和 this .