问题描述
我来自C＃/ F＃/ Haskell，所以我想为我以前使用类型解决的编程问题提供解决方案。
I'm coming from C#/F#/Haskell so I'm trying to come up with solutions to programming problems I'm used to solving with types.
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A类，其中T：C＃中的new（），这主要是因为我可以在某个地方进行new T（）。这会在Dart中创建格式错误的类型错误。是否有合理的惯用方式解决此问题？也许是在工厂？
class A where T : new() in C#, this is mainly so I can do new T() somewhere. This creates a malformed type error in Dart. Is there a reasonably idiomatic way to solve this? Perhaps with factories?
我对mixin进行了一些实验，如果继承的mixin成员的名称发生冲突，则最后一个mixin获胜。因此对于以下内容：
抽象类mixA {void foo（）; }
抽象类mixB {void foo（）; }
C类使用mixA，mixB扩展了对象{}
new C（）。foo（）;
I did some experiments with mixins, in the case of name conflicts for inherited mixin members, the last mixin wins. So for the following:abstract class mixA { void foo(); }abstract class mixB { void foo(); }class C extends Object with mixA, mixB {}new C().foo();
这将最终调用mixB.foo（），而
类C会使用mixB扩展Object，mixA {}
最终将调用mixA.foo（）
要访问隐藏的mixins的成员？
this would end up calling mixB.foo() whereasclass C extends Object with mixB, mixA {}would end up calling mixA.foo()Is there anyway to access the members of the hidden mixins?
假设我将2个mixins与同名字段混合在一起。子类实例在运行时是否具有2个字段（仅1个是不可访问的），或者该对象像字典一样，每个名称只有1个插槽？
Suppose I mix 2 mixins with a field of the same name. Does the subclass instance have 2 fields at runtime (just 1 is inaccessible) or is the object like a dictionary and there is only 1 slot for each name?
推荐答案
1是不可能的。您不能使用通用类型（或与此相关的变量）调用 new 。最常见的解决方法是创建一个分配对象的闭包。
1 is not possible. You can't call new on a generic type (or variable for that matter). The most common workaround is to create a closure that allocates the object instead.
对于2的答案是因为Mixins可以被视为超类： A用B扩展对象，C 基本上等同于：
The answers for 2 fall out of the fact that Mixins can be seen as super-classes: A extends Object with B, C is basically equivalent to:
class B' extends Object { <copy over code inside B> }
class C' extends B' { <copy over code inside C> }
class D extends C' { ... }
考虑到这一点：
- 不。无法访问隐藏的超级元素。
- 是。
小注：<复制X内的代码> 部分并不完全正确，因为那会改变库范围。从概念上讲，mixin B中的代码在声明mixin的库中。
Small note: the <copy over code inside X> part is not completely correct since that would change the library-scope. Code from mixin B is conceptually in the library the mixin is declared in.
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