How remove leading 0 in a string

Question

I get Ids in a String format that contain leading zeroes. I would like to keep this String format but without the leading zeroes. Example:

00 => 0
01 => 1
02 => 2
10 => 10
11 => 11

My current implementation is String id = int.parse(originalId).toString()

Is there a better / efficient / Dart pattern way to achieve this conversion ?

Answer 1

You could use RegExp which may be uglier but faster than double conversion :

"01".replaceAll(new RegExp(r'^0+(?=.)'), '')

´^´ matches the begining of the string

0+ matches one or more 0 character

(=?.) matches a group (()) of any characters except line breaks (.) without including it in the result (=?), this ensures that not the entire string will be matched so that we keep at least one zero if there are only zeroes.

Example :

void main() {

  final List nums = ["00", "01", "02", "10", "11"];
  final RegExp regexp = new RegExp(r'^0+(?=.)');
  for (String s in nums) {
    print("$s => ${s.replaceAll(regexp, '')}");
  }

}

Result :

00 => 0
01 => 1
02 => 2
10 => 10
11 => 11

EDIT : Performance test thanks to your comment

void main() {

  Stopwatch stopwatch = Stopwatch()..start();
  final RegExp reg = RegExp(r'^0+(?=.)');
  for (int i = 0; i < 20000000; i++) {
    '05'.replaceAll(reg, '');
  }
  print('RegExp executed in ${stopwatch.elapsed}');

  stopwatch = Stopwatch()..start();
  for (int i = 0; i < 20000000; i++) {
    int.parse('05').toString();
  }
  print('Double conversion executed in ${stopwatch.elapsed}');

}

Result :

RegExp executed in 0:00:02.912000
Double conversion executed in 0:00:03.216000

The more operations you will do the more it will be efficient compared to double conversion. However RegExp may be slower in a single computation because creating it has a cost, but we speak about a few microseconds... I would say that unless you have tens of thousands of operations just use the more convenient to you.