Java中的通用流利生成器

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呈现我想要的东西.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语法构造它.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 - 沿着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,例如使用递归泛型,但我玩了一段时间,并且无法弄清楚它是如何应用于这个用例的.

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)​​,代表了四种可能的组合T和S类型的定义与否.

它确实有效,但这并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.

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,所以更传统的步骤制作方法就是这样的.

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.我不认为这有很好的办法.

基本思路就是: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实现通用接口.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,并让每个人都通过一种static方法.

否则它的工作方式与您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

这是关于什么的？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伪装成一个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.)

虽然我们不需要复制withNumber,但我只是采取了额外的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.

这里的区别是,我们不保留的类型typed和list如果援引要么自己制定者的第二次.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一样的.

现在我们变得非常新颖.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,但基本思路是:

• 每种方法都以某种方式依赖于在类上声明的类型变量.
• 通过将其类型变量设置为返回类型来"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.

这个例子和前面例子之间的区别在于,如果我们第二次尝试调用setter,我们将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你:

• 由于正当理由,不能再次召集二传手
• 并且可以使用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,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.:)

如果您对这类事情感兴趣,我建议使用注释处理学习代码生成,因为您可以比手动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变得不现实了.