What is the best way to make sure a String can be stored as an int in java

Question

I am making a web-based application and i have text-fields where the values are stored as Strings. The problem is that some of the text-fields are to be parsed into ints and you can store much bigger numbers in Strings than you can in an int. My question is, what is the best way to make sure that the String number can be parsed into an int without erroring out.

Validate the input by parsing with `Integer.parseInt(...)` in a `try/catch(NumberFormatException e)`. — Hovercraft Full Of Eels, Jun 19 '12 at 14:20
How big your number? Is your number it out range of available types in java (Double/Float) ? — Santosh, Jun 19 '12 at 14:21
@HovercraftFullOfEels I don't think the OP is talking about `NumberFormatException`. I think is about the `String` when converted to a number(`Integer`), might be out of range. — Kazekage Gaara, Jun 19 '12 at 14:22
@dystroy if the number is out of range then what happens? Just curious. — Kazekage Gaara, Jun 19 '12 at 14:23
@KazekageGaara: No the OP was not talking directly about using NumberFormatException, but it is a way to solve his main problem. Please look at *many* of the answers below that show specifically how this is done. — Hovercraft Full Of Eels, Jun 19 '12 at 14:29
@user1423793 : Could you please provide your string that you need to parse as int? — Fahim Parkar, Jun 19 '12 at 14:39
Unless you really care about accidental leading or trailing spaces, call String.trim() on the String before calling parseInt() as suggested in all the answers. At least in Java 6, parseInt() throws Exceptions on leading or trailing spaces. — user949300, Jun 19 '12 at 14:43

Michael · Accepted Answer · 2012-06-19T14:32:10.973

11

You can use a try/catch structure for that.

try {
    Integer.parseInt(yourString);
    //new BigInteger(yourString);
    //Use the above if parsing amounts beyond the range of an Integer.
} catch (NumberFormatException e) {
    /* Fix the problem */
}

edited Jun 19 '12 at 14:32

answered Jun 19 '12 at 14:21

Michael

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score 5 · Answer 2 · answered Jun 19 '12 at 14:26

The Integer.parseInt method checks the range as is explicited by the javadoc :

An exception of type NumberFormatException is thrown if any of the following situations occurs:
The first argument is null or is a string of length zero.
The radix is either smaller than Character.MIN_RADIX or larger than Character.MAX_RADIX.
Any character of the string is not a digit of the specified radix, except that the first character may be a minus sign '-' ('\u002D') provided that the string is longer than length 1.
The value represented by the string is not a value of type int.
Examples:
 parseInt("0", 10) returns 0
 parseInt("473", 10) returns 473
 parseInt("-0", 10) returns 0
 parseInt("-FF", 16) returns -255
 parseInt("1100110", 2) returns 102
 parseInt("2147483647", 10) returns 2147483647
 parseInt("-2147483648", 10) returns -2147483648
 parseInt("2147483648", 10) throws a NumberFormatException
 parseInt("99", 8) throws a NumberFormatException
 parseInt("Kona", 10) throws a NumberFormatException
 parseInt("Kona", 27) returns 411787

So the correct way is to try parsing the string :

try {
    Integer.parseInt(str);
} catch (NumberFormatException e) {
    // not an int
}

score 3 · Answer 3 · answered Jun 19 '12 at 14:28

3

Parse the string to a BigInteger instead of a regular Integer. This can hold much higher values.

BigInteger theInteger = new BigInteger(stringToBeParsed);

answered Jun 19 '12 at 14:28

John Snow

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That doesn't seem to answer the question, though. – Denys Séguret Jun 19 '12 at 14:29
@dystroy. Actually I think it does. The way I read the question is that he wants the user to be able to input any number, but don't want the programm to crash due to the number beeing to large – John Snow Jun 19 '12 at 14:31
In fact maybe. Using "int" in the question instead of "integer" would so just be an error. – Denys Séguret Jun 19 '12 at 14:35

score 0 · Answer 4 · answered Jun 19 '12 at 14:21

0

You could run a check in your code:

Convert the String into a long.
Compare the long against the max value for an integer (a constant within the Integer class.)
If the long is greater than it, you know that it cannot be parsed into an int without overflow.
If it is less than or equal to it, convert your long into an int.

answered Jun 19 '12 at 14:21

BlackVegetable

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Why not just convert it straight into an int? – luketorjussen Jun 19 '12 at 14:24
Because I am silly, clearly. (Wouldn't an int possibly overflow to a negative value, or fail? It would require a try/catch block, which admittedly is the better answer.) – BlackVegetable Jun 19 '12 at 14:25

score 0 · Answer 5 · answered Jun 19 '12 at 14:22

0

Always parse the string in try catch block so if any exception or error occurred you know that there is some error in String to int parsing.

answered Jun 19 '12 at 14:22

Pramod Kumar

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score 0 · Answer 6 · answered Jun 19 '12 at 14:29

You could use Apache Commons Lang.

import org.apache.commons.lang.math.NumberUtils;

NumberUtils.toInt( "", 10 );   // returns 10
NumberUtils.toInt( null, 10 ); // returns 10
NumberUtils.toInt( "1", 0 );  // returns 1

The second number is the default if the String is not a numeric value. The first parameter is the String you are trying to convert.

For large numbers I would do the following

BigInteger val = null;
try {
  val = new BigInteger( "1" );
} catch ( NumberFormatException e ) {
  val = BigInteger.ZERO;
}

score 0 · Answer 7 · answered Jun 19 '12 at 14:30

0

what about this ?

BigInteger bigInt = BigInteger(numberAsString);
boolean fitsInInt = ( bigInt.compareTo( BigInteger.valueOf(bigInt.intValue()) ) == 0;

answered Jun 19 '12 at 14:30

Rostislav Matl

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What is the best way to make sure a String can be stored as an int in java

7 Answers7