String.hashcode() Clarification

Question

I was looking at the String.hashcode() function. It's defined as:

This method returns a hash code for this string. The hash code for a String object is computed as:

s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]

Using int arithmetic, where s[i] is the ith character of the string, n is the length of the string, and ^ indicates exponentiation. (The hash value of the empty string is zero.)

The question here is s[0] refers to the 0th character or rather the nth character in the string. But is it the int value of that character that is used to compute the hashcode?

is it the int value of that character that is used to compute the hashcode? Answer is YES. — Santosh
– Santosh, Commented Oct 11, 2013 at 3:47

Thusitha Thilina Dayaratne · Accepted Answer · 2013-10-11 03:45:32Z

2

Every char has a numeric value (ASCII/ UTF-16/ etc...). That is being use in this calculation e.g.

char x = 'A'; System.out.println(x * 2); // output is 130

that is because numeric value of 'A' is 65

answered Oct 11, 2013 at 3:45

Thusitha Thilina Dayaratne

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1 Comment

Ted Hopp Over a year ago

Actually, it's not ASCII, but UTF-16 encoding of Unicode characters.

Evgeniy Dorofeev · Accepted Answer · 2013-10-11 04:21:26Z

The smallest type Java execute arithmetic operations on is int. So s[0]*31^(n-1) is converted to ((int)s[0])*31^(n-1)

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