oop - 广泛使用工厂构造函数是个好主意吗？

Question

我的问题是关于在处理相对复杂的不可变对象(immutable对象)时如何正确使用工厂构造函数。假设我想返回一个更改了某些属性的类的实例。

示例

@immutable
class SomeObject {
  final int id;
  final String name;
  final int upvote;
  final int downvote;
  final int favorite;

  SomeObject({this.id, this.name, this.upvote, this.downvote, this.favorite});

  factory SomeObject.upvoted() {
    return SomeObject(
      upvote: this.upvote + 1 // apparently can't use this keyword here, wrong syntax
    );
  }
  SomeObject.upvoted(SomeObject ref) {
    id = ref.id;
    // cant change an immutable type
  }

  SomeObject upvoted() {
    return SomeObject(
      upvote: upvote + 1,
      //... other properties are null, bad idea?
    );
  }

  SomeObject upvotedWithDefaultConstructorUsingReference(SomeObject ref) {
    // works but seems like an overkill, especially considering more number of properties
    return SomeObject(
      id: ref.id,
      name: ref.name,
      upvote: upvote + 1,
      downvote: ref.downvote,
      favorite: ref.downvote
    );
  }
}

SomeObject.upvoted() 将是同一类的一个实例，但它的 upvoted 属性比引用的属性多 +1。还会有更多类似 downvoted()、withNameChanged() 或 copyWith()。

前两个是构造函数，其他只是返回 SomeObject 类实例的方法。这里应该采用什么方法？当类不可变时如何使用工厂构造函数？我也不确定这 4 个例子的区别。

我已经阅读了 this 的答案问题，但它似乎没有回答我的问题。

Answer 1

看起来你想要一个 copyWith类型模式:

class SomeObject {
  final int id;
  final String name;
  final int upVote;
  final int downVote;
  final int favorite;

  SomeObject({this.id, this.name, this.upVote, this.downVote, this.favorite});

  SomeObject copyWith({
    int id,
    String name,
    int upVote,
    int downVote,
    int favorite,
  }) {
    return SomeObject(
      id: id ?? this.id,
      name: name ?? this.name,
      upVote: upVote ?? this.upVote,
      downVote: downVote ?? this.downVote,
      favorite: favorite ?? this.favorite,
    );
  }
}

你可以随心所欲地调整它: upVoted与 upVote 复制递增，其余部分保持不变，或允许更改。

  SomeObject upVoted() {
    return SomeObject(
      id: id, // no need for 'this' here
      name: name,
      upVote: upVote + 1,
      downVote: downVote,
      favorite: favorite,
    );
  }

将两者结合起来，您可以想出无穷无尽的变化:

  SomeObject upVoted() => copyWith(upVote: upVote + 1);
  SomeObject downVoted() => copyWith(downVote: downVote + 1);
  SomeObject upVoteRetracted() => copyWith(upVote: upVote - 1);

...这让你开始怀疑为什么这个类是不可变的。让它保持和改变它的状态似乎更有意义，而不是制作具有不同值的多个副本。