String age = stdin.readLineSync()!; // first take the input from user in string form
int.parse(age);  // then parse it to integer that's it

 void main(){
  var x = "4";
  int number = int.parse(x);//STRING to INT

  var y = "4.6";
  double doubleNum = double.parse(y);//STRING to DOUBLE

  var z = 55;
  String myStr = z.toString();//INT to STRING
}

int.parse()和double.parse()在无法解析String时抛出错误

每支镖2.6支

int的可选参数onError。不建议使用Parse。因此，应该使用int。tryParse代替。

注意: 这同样适用于double.parse。因此，使用double。tryParse代替。

  /**
   * ...
   *
   * The [onError] parameter is deprecated and will be removed.
   * Instead of `int.parse(string, onError: (string) => ...)`,
   * you should use `int.tryParse(string) ?? (...)`.
   *
   * ...
   */
  external static int parse(String source, {int radix, @deprecated int onError(String source)});

区别在于int。如果源字符串无效，tryParse返回null。

  /**
   * Parse [source] as a, possibly signed, integer literal and return its value.
   *
   * Like [parse] except that this function returns `null` where a
   * similar call to [parse] would throw a [FormatException],
   * and the [source] must still not be `null`.
   */
  external static int tryParse(String source, {int radix});

所以，在你的例子中，它应该是这样的:

// Valid source value
int parsedValue1 = int.tryParse('12345');
print(parsedValue1); // 12345

// Error handling
int parsedValue2 = int.tryParse('');
if (parsedValue2 == null) {
  print(parsedValue2); // null
  //
  // handle the error here ...
  //
}

void main(){
String myString ='111';
int data = int.parse(myString);
print(data);
}

将字符串转换为Int

var myInt = int.parse('12345');
assert(myInt is int);
print(myInt); // 12345
print(myInt.runtimeType);

将字符串转换为Double

var myDouble = double.parse('123.45');
assert(myInt is double);
print(myDouble); // 123.45
print(myDouble.runtimeType);

在DartPad中的例子

你可以用int来解析字符串。解析('你的字符串值');

示例:- int num = int.parse('110011');打印(num);//打印110011;