在Dart 2 int。tryParse可用。

对于无效输入，它返回null而不是抛出。你可以这样使用它:

int val = int.tryParse(text) ?? defaultValue;

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;

你可以这样做，这样更容易转换

示例代码

void main() {
  var myInt = int.parse('12345');
 var number = myInt.toInt();
  print(number); // 12345
  print(number.runtimeType);  // int
  
  
    var myDouble = double.parse('123.45');
  var double_int = myDouble.toDouble();
  print(double_int); // 123.45
   print(double_int.runtimeType);  
}

 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

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