我想把1或32.23这样的字符串解析为整数和双精度数。我怎么能和达特在一起?


当前回答

 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

其他回答

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

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;
 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

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;