Converting 32-bit binary string with Integer.parseInt fails

java integer 32-bit parseint numberformatexception
Aliens picture Aliens · Jan 17, 2012 · Viewed 14.7k times · Source

Why does this part of code fail:

Integer.parseInt("11000000000000000000000000000000",2);

Exception in thread "main" java.lang.NumberFormatException: For input string: "11000000000000000000000000000000"

As far as I understand Integer is a 32 bit value. The number of zeros and ones in the upper code is 32. If there are 31 the code works. Why is that so?

Answer

millimoose picture millimoose · Jan 17, 2012

Your code fails because it tries to parse a number that would require 33 bits to store as a signed integer.

A signed int is a 32 bit value in two's complement representation, where the first bit will indicate the sign of the number, and the remaining 31 bits the value of the number. (-ish.) Java only supports signed integers, and parseInt() and friends aren't supposed to parse two's complement bit patterns – and thus interpret the 1 or (possibly implied) 0 at the 32nd position from the right as the sign. They're meant to support parsing a human-readable reprentation, which is an optional - (or +) for the sign, followed by the absolute value of a number.

In this context, it's a false intuition that leads you to expect the behaviour you describe: if you were parsing any other base besides base 2 (or maybe the other commonly used power-of-two bases), would you expect the first digit of the input to affect the sign? Obviously you wouldn't; having, say, parseInt("2147483648") return -2147483648 by design would be PHP levels of crazy.

Special-casing power-of-two bases also feels odd. Better to have a separate approach to handling bit patterns, for example the one in this answer.

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