Question

我知道有类似的问题。我没有看到我的问题的答案。

我会用一些简化的代码展示我想要的东西。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。我想让它变得优雅。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();

没有不安全的强制转换警告，也不需要预先指定泛型类型（在顶部，构建Builder的位置）。

相反，我们让类型信息明确地流入链中 - 以及withList和withTyped调用。

现在，实现它的最优雅方式是什么？

我知道最常见的技巧，例如使用recursive generics，但我玩了一段时间并且无法弄清楚它是如何适用于这个用例的。

下面是一个平凡的详细解决方案，它在满足所有要求的意义上工作，但代价是冗长 - 它引入了四个构建器（在继承方面无关），代表T和S的四种可能组合是否定义了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); } }种类型。

它确实有效，但这不是一个值得骄傲的版本，如果我们期望更多的通用参数而不仅仅是两个，那就无法维护。

rungeKutta1()
euler1()
rungeKutta2()
euler2()
rungeKutta3()
euler3()

我可以使用更聪明的技术吗？

Answer 1

好的，所以更传统的步骤制作方法就是这样。

不幸的是，因为我们正在混合泛型和非泛型方法，我们必须重新声明许多方法。我不会认为这是一个很好的解决方法。

基本思想就是：定义接口上的每个步骤，然后在私有类上实现它们。我们可以通过继承原始类型来使用通用接口。这很难看，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，因此我们将其设为私有，并让每个人都使用static方法。

否则它与你自己的想法非常相似：

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

这是什么意思？好的，所以这里一般都有问题，就像这样：

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();

如果我们没有创建副本，我们现在123伪装成String。

（如果您只使用构建器作为一组流畅的调用，则不会发生这种情况。）

虽然我们不需要为withNumber制作副本，但我只是采取了额外步骤并使构建器不可变。我们创建的对象比我们要多，但是没有其他好的解决方案。如果每个人都以正确的方式使用构建器，那么我们可以使它变得可变并且return this。

由于我们对新颖的通用解决方案感兴趣，因此这是一个单独的类中的构建器实现。

这里的不同之处在于，如果我们第二次调用它们的任何一个setter，我们就不会保留typed和list的类型。这本身并不是一个缺点，我想它只是不同。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。

现在我们正在真正的小说。这里的想法是我们可以通过导致捕获转换错误来“禁用”每个方法。

解释起来有点复杂，但基本的想法是：

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

此示例与前一个示例的区别在于，如果我们尝试第二次调用setter，我们将收到编译器错误：

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你：

无法再次为合法原因致电设置者
和可以第二次使用null字面值来调用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，关于创建副本和堆污染也是如此。

无论如何，我希望这会给你一些想法让你陷入困境。：）

如果你对这类事情感兴趣，我建议学习code generation with annotation processing，因为你可以比手工编写更容易生成这样的事情。正如我们在评论中所谈到的那样，用手写这样的东西很快就会变得不切实际。

Java中的通用流利生成器

1 个答案: