How to get Integer#parseInt to return a value in case of overload

Question

Trying to run something like below will throw a NumberFormatException:

int i = Integer.parseInt("99999999999999999999999999999999999");

I want this to return Integer.MAX_VALUE instead such that:

int i = foobar("99999999999999999999999999999");
assert(i == Integer.MAX_VALUE);

What is the best way to accomplish this?

What do you want to do if the input is not a valid integer of any size? (eg `foobar("0.5")` or `foobar("pi")`) And what if it's a big negative number? — Dawood ibn Kareem, Nov 21 '19 at 05:09
I do not worry about non-integer values in my case besides overflow — Layman, Nov 21 '19 at 05:11
@ElliottFrisch I have voted to undelete your post. Your approach and the wombat's approach both have value. It would be really good to see them both on this page. — Dawood ibn Kareem, Nov 21 '19 at 05:36
You can't have 99999999999999999999999999999999999 represented as an int. Max value for an int is 2^31-1 — Enrico Giurin, Nov 21 '19 at 05:41

Scary Wombat · Accepted Answer · 2019-11-21T05:08:42.487

2

You could use BigInteger

BigInteger bi = new BigInteger("99999999999999999999999999");
BigInteger mv = new BigInteger(String.valueOf(Integer.MAX_VALUE));
System.out.println(bi.compareTo(mv));

if (bi.compareTo(mv) > 0) 
    return Integer.MAX_VALUE;
else 
    return bi.intValue();

edited Nov 21 '19 at 05:08

answered Nov 21 '19 at 05:05

Scary Wombat

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as per @DawoodsaysreinstateMonica comment below Elliots deleted answer, testing should also be done for large negative numbers too – Scary Wombat Nov 21 '19 at 06:03

Marwa Eldawy · Answer 2 · 2019-11-21T05:12:20.900

1

You can try to parse the number using Integer.parseInt, if it fails, you try to parse it using Long.parseLong. Then you have two ways, if it fails, you try to parse using new BigInteger(stringVal), if it succeeds, you return MAX_INT. Then do the same using BigInteger.

You can't just return max_int if you got a NumberFormatException after attempting to parse it using Integer.parseInt because this exception can be thrown if the text is not a number (or contain characters with numbers).

edited Nov 21 '19 at 05:12

answered Nov 21 '19 at 05:05

Marwa Eldawy

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2
5

`BigInteger.parse` - do you have a link – Scary Wombat Nov 21 '19 at 05:07
You can create a new BigInteger from a string directly. BigInteger b = new BigInteger(string); – Marwa Eldawy Nov 21 '19 at 05:10

score 0 · Answer 3 · answered Nov 21 '19 at 08:07

import java.util.regex.Pattern;

public class ForExample {

  public static void main(String[] args) {
    var strInt = "99999999999999999999999999999999999";

    //
    //
    int i = parseToInt(strInt);
    //
    //

    System.out.println(i == Integer.MAX_VALUE);
  }

  private static final Pattern PLUS_MINUS_ONLY_DIGITS = Pattern.compile("[+-]?\\d+");

  public static int parseToInt(String strInt) {
    try {
      return Integer.parseInt(strInt);
    } catch (NumberFormatException e) {
      if (PLUS_MINUS_ONLY_DIGITS.matcher(strInt).matches()) {
        return strInt.startsWith("-") ? Integer.MIN_VALUE : Integer.MAX_VALUE;
      }
      throw e;
    }
  }

}

How to get Integer#parseInt to return a value in case of overload

3 Answers3