How to parse a String into a number in Dart/Flutter

Converting String into Number is necessary for almost language. Dart is no exception. In this tutorial, we’re gonna look at ways to parse a String into a number (int & double) using parse() and tryParse() method. You also know how to catch Exception for invalid input string, or convert a Hex string to int in Dart/Flutter.

Related Posts:
– Dart/Flutter – Convert/Parse JSON string, array into Object, List
– Dart – Convert Object to Map and Vice Versa

– Dart/Flutter Constructors tutorial with examples
– Dart/Flutter String Methods & Operators tutorial with examples
– Dart/Flutter Future Tutorial with Examples
– Dart/Flutter List Tutorial with Examples
– Dart/Flutter Map Tutorial with Examples

Parse a String into an int in Dart/Flutter

Using parse() method

To parse a string into an int, we use int class parse() method:

var n = int.parse('42');
// 42

The input String could be signed (- or +):

var n1 = int.parse('-42');
// -42

var n2 = int.parse('+42');
// 42

Radix – Hex string to int

Sometimes we have to work with string in radix number format. Dart int parse() method also supports convert string into a number with radix in the range 2..36:

For example, we convert a Hex string into int:

var n_16 = int.parse('FF', radix: 16);
// 255

parse() method and catch Exception

If input string isn’t in a valid integer form, the program throws a FormatException:

var n = int.parse('n42');
/*
Exception has occurred.
FormatException (FormatException: Invalid radix-10 number (at character 1)
n42
^
*/

So to handle this case, we can use Dart try-catch block:

try {
  var n = int.parse('n42');
  print(n);
} on FormatException {
  print('Format error!');
}
// Format error!

int class parse() method also gives us a way to handle FormatException case with onError parameter.

var num4 = int.parse("n42", onError: (source) => -1);
// -1

When the exception is thrown, onError will be called with source as input string. Now we can return an integer value or null… In the example above, the function returns -1 whenever source is in wrong integer literal.

Using tryParse() method

Instead of int.parse(string, onError: (string) => ...), we should use int.tryParse(string):

var n0 = int.tryParse('42');
// 42

var n1 = int.tryParse('-42');
// -42

var n2 = int.tryParse('FF', radix: 16);
// 255

What is the different between parse() and tryParse()?

• parse() throws a FormatException for invalid input string
• tryParse() returns null for invalid input string

tryParse() method and Exception

For tryParse() method, we don’t have to catch Exception, we only need to check null value.
The statement will be simpler:

var n = int.tryParse('n42') ?? -1;
// -1

We can also return a string value while parse() method with onError parameter only returns integer:

var n = int.tryParse('n42') ?? 'Format error!';
// 'Format error!'

Parse a String into a double in Dart/Flutter

Using parse() method

Like integer, string in double literal can be parsed into double using parse() method:

var n1 = double.parse('4.2');
// 4.2

var n2 = double.parse('0.');
// 0.0

var n3 = double.parse('.0');
// 0.0

var n4 = double.parse('-1.e3');
// -1000.0

var n5 = double.parse('123E+3');
// 123000.0

var n6 = double.parse('+.12e-2');
// 0.0012

var n7 = double.parse('+.12e-6');
// 1.2e-7

var n8 = double.parse('-NaN');
// NaN

parse() method and catch Exception

If input string isn’t in a valid double form, the program throws a FormatException:

var n = double.parse('n4.2');
/*
Exception has occurred.
FormatException (FormatException: Invalid double n4.2)
*/

We can use Dart try-catch block to handle this Exception:

try {
  var n = double.parse('n4.2');
  print(n);
} on FormatException {
  print('Format error!');
}
// Format error!

Dart double class parse() method can handle FormatException case with additional onError parameter.

var n = double.parse('n4.2', (source) => -1);
// -1.0

Using tryParse() method

Instead of throwing FormatException for invalid input string, double class tryParse() method returns null:

double n = double.tryParse('n4.2');
// null

Using tryParse() method, we can check null then return specific value.

double n = double.tryParse('n4.2') ?? -1;
// -1.0

Conclusion

We’ve known how to parse a String into a number (int or double) in Dart/Flutter by using parse() or tryParse() method.

Remember that onError parameter is deprecated and will be removed in the future. So tryParse() method could be the better choice.

You can also find many Dart helpful String methods in the post:
Dart/Flutter String Methods & Operators tutorial with examples

Happy learning! See you again!

Further reading

• Dart language tour | Numbers
• Dart Language Tour | String

– Dart/Flutter Constructors tutorial with examples
– Dart/Flutter Future Tutorial with Examples
– Dart/Flutter List Tutorial with Examples
– Dart/Flutter Map Tutorial with Examples