Java fails to parse a hex string as an int

Question

Can someone tell me why this code throws an Exception?

int value = 0xabcdef01;
System.out.println(value);                 // prints -1412567295
String hex = Integer.toHexString(value);
System.out.println(hex);                   // prints abcdef01
// why does this line fail?
Integer.parseInt(hex, 16);

This code throws the following exception:

java.lang.NumberFormatException: For input string: "abcdef01"
  at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
  at java.lang.Integer.parseInt(Integer.java:583)

I'm running on Windows 7 with the following JDK

java version "1.8.0_51"
Java(TM) SE Runtime Environment (build 1.8.0_51-b16)
Java HotSpot(TM) 64-Bit Server VM (build 25.51-b03, mixed mode)

possbile duplicate of stackoverflow.com/questions/11194513/convert-hex-string-to-int — nafas
– nafas, Commented Aug 28, 2015 at 15:46
np mate, btw I just tried that to make sure and by using Long.parseLong(hex,16); works fine. — nafas
– nafas, Commented Aug 28, 2015 at 15:51

Tagir Valeev · Accepted Answer · 2015-08-28 15:55:37Z

1

As you are using Java 8, consider the Integer.parseUnsignedInt method:

Integer.parseUnsignedInt(hex, 16);

answered Aug 28, 2015 at 15:55

Tagir Valeev

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1 Comment

cyber-monk Over a year ago

thanks this solves my problem, I had assumed Integer.parseInt(..., 16) would handle hex which represents a negative int value

Peter Lawrey · Accepted Answer · 2015-08-28 15:47:03Z

1

Perhaps what you wanted was

 int num = (int) Long.parseLong(hex, 16);

The problem is that numbers >= 0x8000_0000 are too large to store in an int

answered Aug 28, 2015 at 15:47

Peter Lawrey

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4 Comments

cyber-monk Over a year ago

My understanding of low level number storage is lacking, but Java int is 32 bits and my hex value is no more than that?

cyber-monk Over a year ago

Also the Long parse routine converts this to a positive long value, when in fact is should be a negative int value

Blake Yarbrough Over a year ago

@cyber-monk actually your number exceeds that by a significant margin, remember that integers can be negative? That means that you actually only get 31 bits for representing your positive numbers. This effectively cuts the maximum value in half.

cyber-monk Over a year ago

@Peter, I had assumed java would take care of the sign for me and convert it into a negative int value

Jacob Eckel · Accepted Answer · 2015-08-28 16:02:09Z

1

Your confusion about integer that doesn't go back to itself has to do with peculiarities of toHexString() which returns "abcdef01" rather than "-543210ff", which really represents your original integer. Run this to see:

  int value = -0x543210ff;
  assert(value == 0xabcdef01);
  assert(value == Integer.parseInt("-543210ff", 16));

edited Aug 28, 2015 at 16:02

answered Aug 28, 2015 at 15:55

Jacob Eckel

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1 Comment

cyber-monk Over a year ago

was unaware java used negative sign indicator for hex strings, thanks

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