Dart-使用地图命名的参数 [英] Dart - named parameters using a Map

问题描述

我想知道是否可以使用地图调用具有名称参数的函数，例如

I would like to know if I can call a function with name parameters using a map e.g.

void main()
{
  Map a = {'m':'done'}; // Map with EXACTLY the same keys as slave named param.
  slave(a);
}
void slave({String m:'not done'}) //Here I should have some type control
{ 
  print(m); //should print done
}

这里的技巧是不使用kwargs，而是使用Map或者，如果您关心类型，那么可以使用一些接口类（就像Json-obj一样），但是让它接受map作为kwars不会更优雅吗？
此外，使用此hack，可选的kwarg可能会很痛苦...
恕我直言，可能的实现方式（如果尚不存在）将是：

the hack here is to not use kwargs but a Map or, if you care about types, some interfaced class (just like Json-obj), but wouldn't be more elegant to just have it accept map as kwars? More, using this hack, optional kwargs would probably become a pain... IMHO a possible implementation, if it does not exist yet, would be something like:

slave(kwargs = a)

例如每个接受命名参数的函数都可以静默地接受一个（Map）kwargs（或其他名称）参数，如果定义了dart，则应在后台执行此逻辑：如果Map中的键正好是非可选键，加上在{}括号中定义的一些可选类型，以及兼容类型为继续。

e.g. Every function that accepts named param could silently accept a (Map) kwargs (or some other name) argument, if defined dart should, under the hood, take care of this logic: if the key in the Map are exactly the non optional ones, plus some of the optional ones, defined in the {} brackets, and of compatible types "go on".

推荐答案

使用 Function.apply 做类似的事情：

main() {
  final a = new Map<Symbol, dynamic>();
  a[const Symbol('m')] = 'done';
  Function.apply(slave, [], a);
}

您还可以提取一个辅助方法来简化代码：

You can also extract an helper method to simplify the code :

main() {
  final a = symbolizeKeys({'m':'done'});
  Function.apply(slave, [], a);
}

Map<Symbol, dynamic> symbolizeKeys(Map<String, dynamic> map){
  final result = new Map<Symbol, dynamic>();
  map.forEach((String k,v) { result[new Symbol(k)] = v; });
  return result;
}

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