int.TryParse returns false when parsing a valid string

Question

I'm trying to check if a string is a valid number or not. But this returns false. Why.

int IsNumber;
var parsed = int.TryParse("9876543210", NumberStyles.Number, null, out IsNumber);

if (parsed)
 {

 }
else
 {

 }

Also tried with CultureInfo.CurrentCulture.NumberFormat, instead of null. When debugging, it never hits inside the if condition.

the maxvalue of a 32-bit integer is 2,147,483,647. use int64/long instead — user1666620
– user1666620, Commented Aug 31, 2015 at 11:08
And if long doesn't cover the range you want, look at BigInteger... — Jon Skeet
– Jon Skeet, Commented Aug 31, 2015 at 11:09
for the future: i guess you could have find that out by using int.Parse and inspect the exception. — nozzleman
– nozzleman, Commented Aug 31, 2015 at 11:09

Kamil Budziewski · Accepted Answer · 2015-08-31 11:09:46Z

12

This should work

long IsNumber;
var parsed = long.TryParse("9876543210", NumberStyles.Number, null, out IsNumber);

if (parsed)
{

}
else
{

}

your code wasn't working because int.Max is 2147483647. If you want to check if all chars in string are digits, you can use Char.IsDigit:

var number = yourString.All(Char.IsDigit);

it will work for numbers bigger than max value of long ( 9223372036854775807)

answered Aug 31, 2015 at 11:09

Kamil Budziewski

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Comments

David Gregor · Accepted Answer · 2015-08-31 11:18:55Z

1

If you are using it for validation (as you describe it) and you don't need to convert it to number type, use regular expression instead. This regular expression gives you information, whether the string contains numbers only

var stringToValidate = "9876543210"
Regex regex = new Regex(@"^\d$");
if (regex.IsMatch(compare))
{ 
    //validation passed
}

answered Aug 31, 2015 at 11:18

David Gregor

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1 Comment

Luaan Over a year ago

You're missing a * (or +) there. Also, careful with \d - it also matches other numerals, like ౫ for example. If you only want "modern european" numbers, [0-9] really works best.

Salah Akbari · Accepted Answer · 2015-08-31 11:59:05Z

1

Use BigInteger.TryParse Method. BigInteger represents an arbitrarily large signed integer.

Add this to using directives at top of page:

using System.Numerics;

Then:

BigInteger IsNumber;
var parsed = BigInteger.TryParse("9876543210", NumberStyles.Integer, null, out IsNumber);

if (parsed)
{

}
else
{

}

edited Aug 31, 2015 at 11:59

answered Aug 31, 2015 at 11:14

Salah Akbari

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