How to convert string to integer in dart

Convert string to integer in Dart:

In this post, we will learn how to convert one string to integer in Dart. Also, we will check different examples with different integers.

Convert string to int in dart:

For converting string to int in dart, we can use either parse or tryParse method. Let’s try them one by one:

String to int using parse:

parse method is defined in integer class in Dart. This is a static method and it is defined as below:

int parse(String strValue, {int? radix})

Here, strValue is the string value of the integer variable and radix is an optional value.

By default, radix is 10 i.e. for decimal conversions. But we can give it a value in range 2..36.

The strValue is the string value that we need to convert to integer. It must not be null. For invalid inputs, it throws FormatException.

Example of parse:

Below are a couple of different use cases of parse:

void main() {
  int firstInt = int.parse('1');
  int secondInt = int.parse('10000');
  int thirdInt = int.parse('-100');
  int fourthInt = int.parse('+100');

  print("firstInt : $firstInt");
  print("secondInt : $secondInt");
  print("thirdInt : $thirdInt");
  print("fourthInt : $fourthInt");
}

It will print:

firstInt : 1
secondInt : 10000
thirdInt : -100
fourthInt : 100

So, as you can see here, we can pasrse strings that defines an integer and even if it has a positive or negative sign at front.

Exceptions in parse:

parse will throw exception for invalid inputs. For example:

void main() {
  try {
    int firstInt = int.parse('99999999999999999999');
  } on FormatException {
    print("FormatException for firstInt");
  }
  try {
    int secondInt = int.parse('2.5');
  } on FormatException {
    print("FormatException for secondInt");
  }
  try {
    int thirdInt = int.parse('2,3');
  } on FormatException {
    print("FormatException for thirdInt");
  }
  try {
    int fourthInt = int.parse('ten');
  } on FormatException {
    print("FormatException for fourthInt");
  }
  try {
    int fifthInt = int.parse(null);
  } on ArgumentError {
    print("Invalid argument exception for fifthInt");
  }
}

All of these cases will throw exception.

For firstInt, we are trying to parse a value more than 2^63 - 1. It should be in the range of -2^63 to 2^63 - 1
For secondInt, parsing doesn’t support any string that contains any separator character.
thirdInt is same as secondInt.
For fourthInt, we are trying to parse a string.
For fifthInt, we are trying to parse null. Note that it throws ArgumentError, not an exception. It throws an error like below:

If you run the above problem, it will print the below output:

FormatException for firstInt
FormatException for secondInt
FormatException for thirdInt
FormatException for fourthInt
Invalid argument exception for fifthInt

Using tryParse:

tryParse is better than parse. It handles the exceptions. For example,

void main() {
    print(int.tryParse('99999999999999999999'));

    print(int.tryParse('2.5'));
  
    print(int.tryParse('2,3'));

    print(int.tryParse('ten'));

    print(int.tryParse(null));
}