Java Integer.parseInt() Not Working for Large Numbers

Question

I have the following simple piece of code which is intended to detect that a given IPv4 address indeed only has numeric values (that is after the dots have been stripped):

import edu.gcc.processing.exceptions.net.IPAddressNumericException;

//Get the IP address
  String address = "239.255.255.255";

//Check to see if this is a number      
  try {
    String IPNumbers = address.replace(".", "");
    Integer.parseInt(IPNumbers);            
  } catch (NumberFormatException e) {
    System.out.print(e.getMessage());
  }

For some reason, the NumberFormatException is fired, and I get this error:

For input string: "239255255255"

Could someone please help me understand this? The parseInt() method works on smaller numbers, such as 127001.

Thank you for your time.

Well, "what is the range of an integer"? (And why might I have asked that question? :-) — , Dec 07 '11 at 21:29
You might want to look into using a regex http://stackoverflow.com/questions/46146/what-are-the-java-regular-expressions-for-matching-ipv4-and-ipv6-strings — Andrew T Finnell, Dec 07 '11 at 21:32

score 11 · Accepted Answer · answered Dec 07 '11 at 21:31

11

try using Long.parseLong(IPNumbers)

answered Dec 07 '11 at 21:31

PTBG

585
2
20
46

score 7 · Answer 2 · answered Dec 07 '11 at 21:31

7

Why not use a regular expression, or break it down into its components by using split(".")?

There are other limitations besides needing to consist solely of digits. For example:

666.666.666.666

This will parse just fine, but is an... unlikely IP.

Breaking it up into its parts lets you determine (a) that it has four parts, and (b) that each part actually makes sense in the context of an IP address.

You could also use an InetAddress implementation, or InetAddressValidator from Apache Commons.

answered Dec 07 '11 at 21:31

Dave Newton

158,873
26
254
302

Also - a very unfortunate one :) – Rion Williams Dec 07 '11 at 21:32
+1 for actually suggesting that the approach of just checking whether it is a number is insufficient, and that switching to a long will not solve this – Robin Dec 07 '11 at 22:03

Rion Williams · Answer 3 · 2011-12-07T21:36:07.780

5

For very large integers you may want to use the BigInteger class (or BigDecimal), as the values may exceed the limits of Integer.

Integer Limits:

Minimum : -2147483648
Maximum :  2147483647

Using BigInteger:

string s = "239255255255";
BigInteger yourNumber = new BigInteger(s);

edited Dec 07 '11 at 21:36

answered Dec 07 '11 at 21:29

Rion Williams

74,820
37
200
327

hmm... What method can I use to convert the string to an integer? – Oliver Spryn Dec 07 '11 at 21:33
There is also a BigDecimal class if you need one, I'll include the necessary documentation in an edit. – Rion Williams Dec 07 '11 at 21:33
239255255255 is not an integer. See http://docs.oracle.com/javase/6/docs/api/java/lang/Integer.html#MAX_VALUE – Steve Kuo Dec 07 '11 at 21:34
@spryno724, An integer will be unable to hold that large of a value, you will have to use BigInteger to store it. – Rion Williams Dec 07 '11 at 21:36

score 3 · Answer 4 · answered Dec 07 '11 at 21:29

3

239255255255 is too big to be held by an Integer. Try using a BigInteger or a BigDecimal.

answered Dec 07 '11 at 21:29

BenCole

2,092
3
17
26

score 3 · Answer 5 · answered Dec 08 '11 at 11:54

Range of an Integer is -2,147,483,648 to 2,147,483,647, the number you have above mentioned is not included in these range, so use long , the range of long primitive type is in between -9,223,372,036,854,775,808 and 9,223,372,036,854,775,807.

So for your case its better to use Long.parseLong() function to convert such large number to long type.

score 1 · Answer 6 · answered Dec 07 '11 at 21:31

1

The logical step up from an integer is a long

answered Dec 07 '11 at 21:31

Ash Burlaczenko

24,778
15
68
99

Java Integer.parseInt() Not Working for Large Numbers

6 Answers6