Java - parse binary to long [duplicate]

Question

I have a binary represenation of a number and want to convert it to long (I have Java 8)

public class TestLongs {
public static void main(String[] args){
    String a = Long.toBinaryString(Long.parseLong("-1")); // 1111111111111111111111111111111111111111111111111111111111111111
    System.out.println(a);
    System.out.println(Long.parseLong(a, 2));// ??? but Long.parseUnsignedLong(a, 2) works
}

}

This code results in Exception in thread "main" java.lang.NumberFormatException: For input string: "1111111111111111111111111111111111111111111111111111111111111111" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) 1111111111111111111111111111111111111111111111111111111111111111 at java.lang.Long.parseLong(Long.java:592) What is wrong here? Why Long.parseLong(a, 2) doesn't work?

Eran · Accepted Answer · 2016-10-18 11:01:12Z

5

Long.parseLong() doesn't treat the first '1' character as a sign bit, so the number is parsed as 2^64-1, which is too large for long. Long.parseLong() expects input Strings that represent negative numbers to start with '-'.

In order for Long.parseLong(str,2) To return -1, you should pass to it a String that start with '-' and ends with the binary representation of 1 - i.e. Long.parseLong("-1",2).

edited Oct 18, 2016 at 11:01

answered Oct 18, 2016 at 10:53

Eran

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11 Comments

android_dev Over a year ago

then what is difference between Long.parseLong and Long.parseUnsignedLong? does "Unsigned" refer to binary represenation or long value??

Peter Lawrey Over a year ago

@android_dev parseUnsignedLong expects to treat the long value as if it was unsigned.

Eran Over a year ago

Long.parseUnsignedLong treats the output long as an unsigned value, so when you treat a long value as unsigned, the max possible value becomes 2^64-1 (represented by 64 1 bits), which is why a String of 64 1s is acceptable. @android_dev

Peter Lawrey Over a year ago

@android_dev parseUnsignedInt parses large unsigned values. long is still signed and no method changes that. You can treat as if it was unsigned of course.

Eran Over a year ago

@android_dev long is still signed, so when you print Long.parseUnsignedLong("11111111111111111‌11111111111111111111‌11111111111111111111‌1111111", 2)) you see -1. In order to see this number as an unsigned long, you need to print it with the static method Long.toUnsignedString().

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ΦXocę 웃 Пepeúpa ツ · Accepted Answer · 2016-10-18 10:58:22Z

3

Eran is right, and the answer to your question is:

System.out.println(new BigInteger(a, 2).longValue());

answered Oct 18, 2016 at 10:58

ΦXocę 웃 Пepeúpa ツ

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