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Asked: May 10, 20262026-05-10T22:08:57+00:00 2026-05-10T22:08:57+00:00

Per the Java documentation, the hash code for a String object is computed as:

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Per the Java documentation, 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.

Why is 31 used as a multiplier?

I understand that the multiplier should be a relatively large prime number. So why not 29, or 37, or even 97?

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  1. 2026-05-10T22:08:57+00:00Added an answer on May 10, 2026 at 10:08 pm

    According to Joshua Bloch’s Effective Java, Second Edition (a book that can’t be recommended enough, and which I bought thanks to continual mentions on Stack Overflow):

    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.

    (from Chapter 3, Item 9: Always override hashCode when you override equals, page 48)

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