有效 Java hashCode() 实现中的位移位

我想知道是否有人可以详细解释一下

(int)(l ^ (l >>> 32));

在以下 hashcode 实现中执行操作（由 eclipse 生成，但与有效 Java 相同）：

private int i;
private char c; 
private boolean b;
private short s;
private long l;
private double d;
private float f;

@Override
public int hashCode() {
    final int prime = 31;
    int result = 1;
    result = prime * result + i;
    result = prime * result + s;
    result = prime * result + (b ? 1231 : 1237);
    result = prime * result + c;
    long t = Double.doubleToLongBits(d);
    result = prime * result + (int) (t ^ (t >>> 32));
    result = prime * result + Float.floatToIntBits(f);
    result = prime * result + (int) (l ^ (l >>> 32));
    return result;
}

Thanks!

基本上，它将 long 的高 32 位与低 32 位进行异或。这是一个分解版本：

// Unsigned shift by 32 bits, so top 32 bits of topBits will be 0,
// bottom 32 bits of topBits will be the top 32 bits of l
long topBits = l >>> 32;

// XOR topBits with l; the top 32 bits will effectively be left
// alone, but that doesn't matter because of the next step. The
// bottom 32 bits will be the XOR of the top and bottom 32 bits of l
long xor = l ^ topBits;

// Convert the long to an int - this basically ditches the top 32 bits
int hash = (int) xor;

To answer your comment: you have a long value which has to be converted into an int to be part of the hash (the result has to only be 32 bits). How are you going to do that? You could just take the bottom 32 bits - but then that means changes in only the top 32 bits would be ignored, which wouldn't make it a very good hash. This way, a change in a single bit of input always results in a change of a single bit of the hash. Admittedly you can still get collisions easily - change both bits 7 and 39, for example, or any other pair of bits 32 positions apart - but that's bound to be the case, given that you're going from 2⁶⁴ possible values to 2³².