Java中String类的hashCode()方法中的魔术数：3

Java中String类的hashCode()方法实现如下：

public int hashCode() {
    int h = hash;
    if (h == 0 && value.length > 0) {
        char val[] = value;

        for (int i = 0; i < value.length; i++) {
            h = 31 * h + val[i];
        }
        hash = h;
    }
    return h;
}

可以看到，循环中的每一步都对上一步的结果乘以一个系数31，那么，这个31又是怎么来的呢？
在《Effective Java》第二版的 Item 9: Always override hashCode when you override equals 中我们找到了答案：

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.

归纳如下：

• 奇数，乘法运算溢出时不丢失信息
• 质数，传统做法
• 可优化 31 * i == (i << 5) - i