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private named parameter in dart

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flutter

dart

I am new to Dart. Let's say I have a final List <Character> _characterList; which should be private. But how do I use CharacterList ({Key key, this._characterList}): super (key: key); if the named parameter cannot start with _?

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Sam324 Avatar asked Oct 05 '20 20:10

Sam324


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2 Answers

The name of the parameter is independent of the name of the member. Constructors offer the this.name syntactic sugar for convenience if the names happen to be the same, but you don't have to use that. You could give the parameter its own name and explicitly initialize the member variable separately:

CharacterList ({Key key, List<Character> characterList})
  : _characterList = characterList,
    super(key: key);

final List<Character> _characterList;
like image 186
jamesdlin Avatar answered Nov 03 '22 05:11

jamesdlin


It is Dart limitation

Named optional parameters can't start with an underscore

You can use a constructor with optional positional parameter instead

CharacterList (Key key, [this._characterList]): super (key: key)
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Pavel Shastov Avatar answered Nov 03 '22 04:11

Pavel Shastov