Integer.parseInt() doesn't parse large negative numbers

Question

Integer.parseInt() doesn't parse large negative numbers

Why is NumberFormatException is thrown when i try Integer.parseInt("80000010", 16)?? That IS a 32-bit number, which is the size of java's int.

EDIT: The best part is this...

int z = 0x80000010;
System.err.println("equal to " + z);

Prints -2147483632 which is indeed 0x80000010 according to my calculator ;)

java

asked on Stack Overflow Mar 30, 2011 by

Saideira • edited Mar 30, 2011 by

Saideira

6 Answers

Because parseInt "parses the string argument as a signed integer", as specified in the API documentation. The value 80000010 with radix 16 is outside the valid range for a signed 32 byte value.

answered on Stack Overflow Mar 30, 2011 by

jarnbjo

You can do

int value = (int) Long.parseLong("80000010", 16)

Update:

With Java 8 update (2014) you can write

int value = Integer.parseUnsignedInt("80000010", 16);

answered on Stack Overflow Mar 30, 2011 by

Peter Lawrey • edited Jul 21, 2014 by

kajacx

80,00,00,10₁₆ = 2,147,483,664₁₀

Whilst a Java integer has a range of -2,147,483,648₁₀ to 2,147,483,647₁₀.

answered on Stack Overflow Mar 30, 2011 by

Paul Ruane

For those stuck with Java <= 7, Guava provides a utility for parsing ints as unsigned:

UnsignedInts.parseUnsignedInt("ffffffff", 16);
> -1

answered on Stack Overflow Jun 8, 2015 by

Natix

It is because the first parameter is signed integer, so for negative numbers you have to give minus sign explicitly. And in your case you have big unsgined number outside the integer range.

answered on Stack Overflow Mar 30, 2011 by

xappymah • edited Mar 30, 2011 by

xappymah

You're parsing it with base 16. So it is greater than the maximum for integer.

answered on Stack Overflow Mar 30, 2011 by

Riccardo Cossu

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