parse method Null safety

num parse(
  1. String input,
  2. [@deprecated num onError(
    1. String input

Parses a string containing a number literal into a number.

The method first tries to read the input as integer (similar to int.parse without a radix). If that fails, it tries to parse the input as a double (similar to double.parse). If that fails, too, it invokes onError with input, and the result of that invocation becomes the result of calling parse.

If no onError is supplied, it defaults to a function that throws a FormatException.

For any number n, this function satisfies identical(n, num.parse(n.toString())) (except when n is a NaN double with a payload).

The onError parameter is deprecated and will be removed. Instead of num.parse(string, (string) { ... }), you should use num.tryParse(string) ?? (...).


static num parse(String input, [@deprecated num onError(String input)?]) {
  num? result = tryParse(input);
  if (result != null) return result;
  if (onError == null) throw FormatException(input);
  return onError(input);