java - Java 中的通用 fluent Builder

Question

我知道有类似的问题。不过，我还没有看到我的问题的答案。

我将使用一些简化的代码来展示我想要的内容。假设我有一个复杂的对象，它的一些值是通用的:

public static class SomeObject<T, S> {
    public int number;
    public T singleGeneric;
    public List<S> listGeneric;

    public SomeObject(int number, T singleGeneric, List<S> listGeneric) {
        this.number = number;
        this.singleGeneric = singleGeneric;
        this.listGeneric = listGeneric;
    }
}

我想用流畅的 Builder 语法构建它。不过，我想让它优雅。我希望它能像那样工作:

SomeObject<String, Integer> works = new Builder() // not generic yet!
    .withNumber(4) 

    // and only here we get "lifted"; 
    // since now it's set on the Integer type for the list
    .withList(new ArrayList<Integer>()) 

    // and the decision to go with String type for the single value
    // is made here:
    .withTyped("something") 

    // we've gathered all the type info along the way
    .create();

没有不安全的转换警告，也不需要预先指定泛型类型(在顶部，构造 Builder 的地方)。

相反，我们让类型信息显式流入，在链的更下方——连同 withList和 withTyped电话。

现在，实现它的最优雅的方法是什么？

我知道最常见的技巧，例如使用 recursive generics ，但我玩了一会儿，无法弄清楚它如何适用于这个用例。

下面是一个普通的冗长解决方案，它可以满足所有要求，但代价是过于冗长——它引入了四个构建器(在继承方面不相关)，代表了 T 的四种可能组合。和 S类型是否被定义。

它确实有效，但是这不是一个值得骄傲的版本，如果我们期望的通用参数不仅仅是两个，那么它就无法维护。

public static class Builder  {
    private int number;

    public Builder withNumber(int number) {
        this.number = number;
        return this;
    }

    public <T> TypedBuilder<T> withTyped(T t) {
        return new TypedBuilder<T>()
                .withNumber(this.number)
                .withTyped(t);
    }

    public <S> TypedListBuilder<S> withList(List<S> list) {
        return new TypedListBuilder<S>()
                .withNumber(number)
                .withList(list);
    }
}

public static class TypedListBuilder<S> {
    private int number;
    private List<S> list;

    public TypedListBuilder<S> withList(List<S> list) {
        this.list = list;
        return this;
    }

    public <T> TypedBothBuilder<T, S> withTyped(T t) {
        return new TypedBothBuilder<T, S>()
                .withList(list)
                .withNumber(number)
                .withTyped(t);
    }

    public TypedListBuilder<S> withNumber(int number) {
        this.number = number;
        return this;
    }
}

public static class TypedBothBuilder<T, S> {
    private int number;
    private List<S> list;
    private T typed;

    public TypedBothBuilder<T, S> withList(List<S> list) {
        this.list = list;
        return this;
    }

    public TypedBothBuilder<T, S> withTyped(T t) {
        this.typed = t;
        return this;
    }

    public TypedBothBuilder<T, S> withNumber(int number) {
        this.number = number;
        return this;
    }

    public SomeObject<T, S> create() {
        return new SomeObject<>(number, typed, list);
    }
}

public static class TypedBuilder<T> {
    private int number;
    private T typed;

    private Builder builder = new Builder();

    public TypedBuilder<T> withNumber(int value) {
        this.number = value;
        return this;
    }

    public TypedBuilder<T> withTyped(T t) {
        typed = t;
        return this;
    }

    public <S> TypedBothBuilder<T, S> withList(List<S> list) {
        return new TypedBothBuilder<T, S>()
                .withNumber(number)
                .withTyped(typed)
                .withList(list);
    }
}

我可以应用更聪明的技术吗？

Answer 1

好的，所以更传统的步骤生成器方法应该是这样的。

不幸的是，因为我们混合了泛型和非泛型方法，我们不得不重新声明很多方法。我认为没有解决这个问题的好方法。

基本思想就是:在接口(interface)上定义每个步骤，然后在私有(private)类上实现它们。我们可以通过继承原始类型来使用通用接口(interface)来做到这一点。这很丑陋，但它确实有效。

public interface NumberStep {
    NumberStep withNumber(int number);
}
public interface NeitherDoneStep extends NumberStep {
    @Override NeitherDoneStep withNumber(int number);
    <T> TypeDoneStep<T> withTyped(T type);
    <S> ListDoneStep<S> withList(List<S> list);
}
public interface TypeDoneStep<T> extends NumberStep {
    @Override TypeDoneStep<T> withNumber(int number);
    TypeDoneStep<T> withTyped(T type);
    <S> BothDoneStep<T, S> withList(List<S> list);
}
public interface ListDoneStep<S> extends NumberStep {
    @Override ListDoneStep<S> withNumber(int number);
    <T> BothDoneStep<T, S> withTyped(T type);
    ListDoneStep<S> withList(List<S> list);
}
public interface BothDoneStep<T, S> extends NumberStep {
    @Override BothDoneStep<T, S> withNumber(int number);
    BothDoneStep<T, S> withTyped(T type);
    BothDoneStep<T, S> withList(List<S> list);
    SomeObject<T, S> create();
}
@SuppressWarnings({"rawtypes","unchecked"})
private static final class BuilderImpl implements NeitherDoneStep, TypeDoneStep, ListDoneStep, BothDoneStep {
    private final int number;
    private final Object typed;
    private final List list;

    private BuilderImpl(int number, Object typed, List list) {
        this.number = number;
        this.typed  = typed;
        this.list   = list;
    }

    @Override
    public BuilderImpl withNumber(int number) {
        return new BuilderImpl(number, this.typed, this.list);
    }

    @Override
    public BuilderImpl withTyped(Object typed) {
        // we could return 'this' at the risk of heap pollution
        return new BuilderImpl(this.number, typed, this.list);
    }

    @Override
    public BuilderImpl withList(List list) {
        // we could return 'this' at the risk of heap pollution
        return new BuilderImpl(this.number, this.typed, list);
    }

    @Override
    public SomeObject create() {
        return new SomeObject(number, typed, list);
    }
}

// static factory
public static NeitherDoneStep builder() {
    return new BuilderImpl(0, null, null);
}

由于我们不希望人们访问丑陋的实现，我们将其设为私有(private)并让每个人都通过 static 方法。

否则它的工作原理与您自己的想法几乎相同:

SomeObject<String, Integer> works =
    SomeObject.builder()
        .withNumber(4)
        .withList(new ArrayList<Integer>())
        .withTyped("something")
        .create();

// we could return 'this' at the risk of heap pollution

这是怎么回事？好的，所以这里有一个一般性的问题，它是这样的:

NeitherDoneStep step = SomeObject.builder();
BothDoneStep<String, Integer> both =
    step.withTyped("abc")
        .withList(Arrays.asList(123));
// setting 'typed' to an Integer when
// we already set it to a String
step.withTyped(123);
SomeObject<String, Integer> oops = both.create();

如果我们不创建副本，我们现在会将 123 伪装成 String。

(如果您仅将构建器用作流畅的调用集，则不会发生这种情况。)

虽然我们不需要为 withNumber 制作副本，但我只是做了额外的一步，让构建器不可变。我们正在创建比我们必须创建的更多的对象，但实际上并没有另一个好的解决方案。如果每个人都打算以正确的方式使用构建器，那么我们可以使它可变并返回它。

---

由于我们对新颖的通用解决方案感兴趣，这里是单个类中的构建器实现。

这里的区别在于，如果我们第二次调用它们的任何一个 setter ，我们不会保留 typed 和 list 的类型。这本身并不是真正的缺点，我猜只是不同而已。这意味着我们可以这样做:

SomeObject<Long, String> =
    SomeObject.builder()
        .withType( new Integer(1) )
        .withList( Arrays.asList("abc","def") )
        .withType( new Long(1L) ) // <-- changing T here
        .create();

public static class OneBuilder<T, S> {
    private final int number;
    private final T typed;
    private final List<S> list;

    private OneBuilder(int number, T typed, List<S> list) {
        this.number = number;
        this.typed  = typed;
        this.list   = list;
    }

    public OneBuilder<T, S> withNumber(int number) {
        return new OneBuilder<T, S>(number, this.typed, this.list);
    }

    public <TR> OneBuilder<TR, S> withTyped(TR typed) {
        // we could return 'this' at the risk of heap pollution
        return new OneBuilder<TR, S>(this.number, typed, this.list);
    }

    public <SR> OneBuilder<T, SR> withList(List<SR> list) {
        // we could return 'this' at the risk of heap pollution
        return new OneBuilder<T, SR>(this.number, this.typed, list);
    }

    public SomeObject<T, S> create() {
        return new SomeObject<T, S>(number, typed, list);
    }
}

// As a side note,
// we could return e.g. <?, ?> here if we wanted to restrict
// the return type of create() in the case that somebody
// calls it immediately.
// The type arguments we specify here are just whatever
// we want create() to return before withTyped(...) and
// withList(...) are each called at least once.
public static OneBuilder<Object, Object> builder() {
    return new OneBuilder<Object, Object>(0, null, null);
}

创建副本和堆污染也是如此。

---

现在我们变得真正新颖了。这里的想法是我们可以通过引起捕获转换错误来“禁用”每个方法。

解释起来有点复杂，但基本思路是:

• 每个方法都以某种方式依赖于在类上声明的类型变量。
• 通过将其返回类型将该类型变量设置为 ? 来“禁用”该方法。
• 如果我们尝试对该返回值调用该方法，则会导致捕获转换错误。

这个例子和前面例子的区别在于，如果我们尝试第二次调用一个setter，我们会得到一个编译器错误:

SomeObject<Long, String> =
    SomeObject.builder()
        .withType( new Integer(1) )
        .withList( Arrays.asList("abc","def") )
        .withType( new Long(1L) ) // <-- compiler error here
        .create();

因此，我们只能调用每个 setter 一次。

这里的两个主要缺点是您:

• 由于正当原因不能再次调用 setter
• 和可以使用null 文字第二次调用setter。

我认为这是一个非常有趣的概念验证，即使它有点不切实际。

public static class OneBuilder<T, S, TCAP, SCAP> {
    private final int number;
    private final T typed;
    private final List<S> list;

    private OneBuilder(int number, T typed, List<S> list) {
        this.number = number;
        this.typed  = typed;
        this.list   = list;
    }

    public OneBuilder<T, S, TCAP, SCAP> withNumber(int number) {
        return new OneBuilder<T, S, TCAP, SCAP>(number, this.typed, this.list);
    }

    public <TR extends TCAP> OneBuilder<TR, S, ?, SCAP> withTyped(TR typed) {
        // we could return 'this' at the risk of heap pollution
        return new OneBuilder<TR, S, TCAP, SCAP>(this.number, typed, this.list);
    }

    public <SR extends SCAP> OneBuilder<T, SR, TCAP, ?> withList(List<SR> list) {
        // we could return 'this' at the risk of heap pollution
        return new OneBuilder<T, SR, TCAP, SCAP>(this.number, this.typed, list);
    }

    public SomeObject<T, S> create() {
        return new SomeObject<T, S>(number, typed, list);
    }
}

// Same thing as the previous example,
// we could return <?, ?, Object, Object> if we wanted
// to restrict the return type of create() in the case
// that someone called it immediately.
// (The type arguments to TCAP and SCAP should stay
// Object because they are the initial bound of TR and SR.)
public static OneBuilder<Object, Object, Object, Object> builder() {
    return new OneBuilder<Object, Object, Object, Object>(0, null, null);
}

同样，创建副本和堆污染也是如此。

---

无论如何，我希望这能给您一些想法，让您全神贯注。 :)

如果你一般对这类事情感兴趣，我建议学习 code generation with annotation processing , 因为你可以生成这样的东西比手写它们容易得多。正如我们在评论中谈到的那样，手写这样的东西很快就会变得不现实。