Converting 32-bit binary string with Integer.parse

问题:

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?

回答1:

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.

回答2:

According to the docs, the max value of an Integer is 2^31-1. Which, in binary is:

1111111111111111111111111111111

In other words, 31 1's in a row.

回答3:

This is because for Integer.parseInt "11000000000000000000000000000000" is not a two's complement representation of -1073741824 but a positive value 3221225472 which does not fit into int values range -2147483648 to 2147483647. But we can parse two's complement binary string representation with BigInteger:

int i = new BigInteger("11000000000000000000000000000000", 2).intValue()

this gives expected -1073741824 result

回答4:

Even though your string, "11.....lots of zeros" is a legal binary representation of a negative integer, Integer.parseInt() fails on it. I consider this a bug.

Adding a little levity, since on rereading this post it sounds too pedantic, I understand that Oracle probably doesn't care much whether I think this is a bug or not. :-)

You can try:

   long avoidOverflows = Long.parseLong("11000000000000000000000000000000",2);
   int thisShouldBeANegativeNumber = (int)avoidOverflows);
   System.out.println(avoidOverflows + " -> " + thisShouldBeANegativeNumber);

you should see
3221225472 -> -1073741824

You sometimes have to do this with Colors depending on how they are stored as text.

BTW, exact thing can happen if you are parsing a Hex representation and you are parsing a negative number like "88888888". You need to use Long.parseLong() then convert.