Say I have
static System::Numerics::BigInteger MinimoNumero; int16_t Uno = 1;
static System::Numerics::BigInteger MaximoNumero;
static const std::string MaximoNumeroString = "91389681247993671255432112333333333333333333333333333333333333333333333333333333333333333333333333333333";
MaximoNumero = System::Numerics::BigInteger::Parse(marshal_as<String^>(MaximoNumeroString));
MinimoNumero = System::Numerics::BigInteger::Parse("1");
How can I SUM 1 to MaximoNumero so I want a result as BigInteger as 91389681247993671255432112333333333333333333333333333333333333333333333333333333333333333333333333333334
If I use
System::Numerics::BigInteger NUM = MinimoNumero + MaximoNumero;
then I got error "more than one operator "+" matches these operands.."
You can't (generally) use the plain arithmetic operators on the BigInteger type. Instead, call the relevant member function(s) of the BigInteger type – in this case, the Add function:
System::Numerics::BigInteger NUM = System::Numerics::BigInteger::Add(MinimoNumero, MaximoNumero);
I'd like to convert a int digit to a char (no to convert from ascii code). I'd like to convert it from int type to char type. (Example: from 5 to '5') Is it possible?
Use:
int x = 5; char c = '0' + x; // c == '5'
So if I have a String
char string[4];
string = "A10";
How Can I get 10 as an Integer.
I tried getting 10 by itself using this but it didn't work.
char string[4];
char string2[2];
string = "A10";
string2[0] = string[1];
string2[1] = string[2];
I don't need to worry about the A, I know how to get that, I need to be able to get the 10 as an integer.
In java you can use the following code to convert string to integer,
int i=Integer.parseInt(s2);
for more information view this website,
https://www.javatpoint.com/java-string-to-int
I would like to parse strings like 1 or 32.23 into integers and doubles. How can I do this with Dart?
You can parse a string into an integer with int.parse(). For example:
var myInt = int.parse('12345');
assert(myInt is int);
print(myInt); // 12345
Note that int.parse() accepts 0x prefixed strings. Otherwise the input is treated as base-10.
You can parse a string into a double with double.parse(). For example:
var myDouble = double.parse('123.45');
assert(myDouble is double);
print(myDouble); // 123.45
parse() will throw FormatException if it cannot parse the input.
In Dart 2 int.tryParse is available.
It returns null for invalid inputs instead of throwing. You can use it like this:
int val = int.tryParse(text) ?? defaultValue;
Convert String to Int
var myInt = int.parse('12345');
assert(myInt is int);
print(myInt); // 12345
print(myInt.runtimeType);
Convert String to Double
var myDouble = double.parse('123.45');
assert(myInt is double);
print(myDouble); // 123.45
print(myDouble.runtimeType);
Example in DartPad
As per dart 2.6
The optional onError parameter of int.parse is deprecated. Therefore, you should use int.tryParse instead.
Note:
The same applies to double.parse. Therefore, use double.tryParse instead.
/**
* ...
*
* 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)});
The difference is that int.tryParse returns null if the source string is invalid.
/**
* 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});
So, in your case it should look like:
// 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(){
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() and double.parse() can throw an error when it couldn't parse the String
Above solutions will not work for String like:
String str = '123 km';
So, the answer in a single line, that works in every situation for me will be:
int r = int.tryParse(str.replaceAll(RegExp(r'[^0-9]'), '')) ?? defaultValue;
or
int? r = int.tryParse(str.replaceAll(RegExp(r'[^0-9]'), ''));
But be warned that it will not work for the below kind of string
String problemString = 'I am a fraction 123.45';
String moreProblem = '20 and 30 is friend';
If you want to extract double which will work in every kind then use:
double d = double.tryParse(str.replaceAll(RegExp(r'[^0-9\.]'), '')) ?? defaultValue;
or
double? d = double.tryParse(str.replaceAll(RegExp(r'[^0-9\.]'), ''));
This will work for problemString but not for moreProblem.
you can parse string with int.parse('your string value');.
Example:- int num = int.parse('110011'); print(num); // prints 110011 ;
If you don't know whether your type is string or int you can do like this:
int parseInt(dynamic s){
if(s.runtimeType==String) return int.parse(s);
return s as int;
}
For double:
double parseDouble(dynamic s){
if(s.runtimeType==String) return double.parse(s);
return s as double;
}
Therefore you can do parseInt('1') or parseInt(1)
void main(){
String myString ='111';
int data = int.parse(myString);
print(data);
}
String age = stdin.readLineSync()!; // first take the input from user in string form
int.parse(age); // then parse it to integer that's it
You can do this for easy conversion like this
Example Code Here
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);
}
How do you convert an unsigned long to a String ^?
I've tried
String ^ mystring = marshal_as<String ^>(myunsignedlong)
but it doesnt work!
"unsigned long" is an alias for System::UInt32 in a C++/CLI program. Use its ToString() method:
unsigned long value = 42;
String^ txt = value.ToString();
or use its overloads to use a non-default format or culture. Or use String::Format() for many more composite formatting options:
String^ txt = String::Format("The value is {0}", value);