dart-从泛型扩展中提取静态方法 [英] dart - pulling static methods from generic extensions

问题描述

这是我在dart中实现的单例模式代码：

Here's the singleton pattern code I was implementing in dart:

void main() {
  var func = new Functioner();
  print("The output is: ${func.divide()(8) }");
}

class Functioner{
Function divide(){
  Function letUsDivide = (int x)=>x~/4;
  return letUsDivide;
  }
  
  static Functioner _mThis = null;
  
  Functioner(){
    _mThis = this;
  }
  
  static Functioner getInstance(){
    if(_mThis == null)
      _mThis = new Functioner();
    return _mThis;
  }
  
  
  int divideInts(int a, int b){
    return (a/b).toInt();
  }
  
  
  
}

class MathGen<T extends Functioner>{
  MathGen();
  
  int divide(){
    Functioner mVal = T;
    double a = 3.5;
    
        
    return T.getInstance().divideInts(3,54); #error
  }
  
}
  

由于允许使用dart的仿制药它们是特定类型的子类，我认为可以使用静态字段和方法，但是我错了。如在 MathGen 类中标记为 error 的行中所述：方法getInstance（）是' t为类型'Type'定义，但是

Since dart's generics allowed making them a subclass of a particular type, I thought static fields and methods would be allowed, but I was wrong. As marked inside the MathGen class with error line says: the method getInstance() isn't defined for the type 'Type' , but

1. 这种语法不应该 T扩展Functioner 表示 T 是否至少是
   Functioner 类型？


2. 然后为什么 getInstance（）不可用


3. 以及如何投射 T 到 Functioner ？ （从我发现
   飞镖似乎没有显式的转换机制？）

1. shouldn't this syntax T extends Functioner mean that T is atleast a Functioner type?
2. then why is getInstance() not available
3. and how do I cast T to a Functioner? (From what I have found dart seems to not have a explicit casting mechanism?)

推荐答案

您可以执行以下操作：

   Functioner getInstance(){
    if(_mThis == null)
      _mThis = new Functioner();
    return _mThis;
  }

使 getInstance（）成为实例方法，然后这样称呼它：

make getInstance() an instance method, then call it like this:

class MathGen<T extends Functioner>{
  MathGen();
  
  T widget;
  
  int divide(){
    double a = 3.5;
    
        
    return widget.getInstance().divideInts(3,54);
  }
  
}

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