如何手动计算字符串的哈希码? [英] How to calculate the hash code of a string by hand?

问题描述

我想知道如何手动计算给定字符串的哈希码.我了解在Java中，您可以执行以下操作:

I was wondering how to calculate the hash code for a given string by hand. I understand that in Java, you can do something like:

String me = "What you say what you say what?";  
long whatever = me.hashCode();

这一切都很好，但我想知道如何手工完成.我知道给定的用于计算字符串的哈希码的公式类似于:

That's all good and dandy, but I was wondering how to do it by hand. I know the given formula for calculating the hash code of a string is something like:

S0 X 31 ^ (n-1) + S1 X 31 ^ (n-2) + .... + S(n-2) X 31 + S(n-1)  

其中S表示字符串中的字符，n是字符串的长度.然后使用16位unicode，来自字符串me的第一个字符将计算为:

Where S indicates the character in the string, and n is the length of the string. Using 16 bit unicode then, the first character from string me would be computed as:

87 X (31 ^ 34)

但是，这会产生惊人的数字.我无法想象像这样将所有字符加在一起.那么，为了计算最低阶的32位结果，我该怎么办?上方的任何东西等于-957986661，我不是怎么计算呢?

However, that creates an insanely large number. I can't imagine adding all the characters together like that. So, in order to calculate the lowest-order 32 bits result then, what would I do? Long whatever from above equals -957986661 and I'm not how to calculate that?

推荐答案

看看java.lang.String的源代码.

/**
 * Returns a hash code for this string. The hash code for a
 * <code>String</code> object is computed as
 * <blockquote><pre>
 * s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]
 * </pre></blockquote>
 * using <code>int</code> arithmetic, where <code>s[i]</code> is the
 * <i>i</i>th character of the string, <code>n</code> is the length of
 * the string, and <code>^</code> indicates exponentiation.
 * (The hash value of the empty string is zero.)
 *
 * @return  a hash code value for this object.
 */
public int hashCode() {
    int h = hash;
    int len = count;
    if (h == 0 && len > 0) {
        int off = offset;
        char val[] = value;
        for (int i = 0; i < len; i++) {
            h = 31*h + val[off++];
        }
        hash = h;
    }
    return h;
}

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