假设我有一个接口Foo。它由具体实现CompositeFooFooAFooBFooC实现。此外,CompositeFoo如下所示:

public class CompositeFoo implements Foo {
    @Inject public CompositeFoo(List<? extends Foo> elements);
}


我希望在Guice PrivateModule中将Foo绑定到CompositeFoo,列表是FooA,后跟FooBFooC。 (这必须是列表,因为顺序很重要;这排除了多重绑定作为解决方案。)

问题是我看到涉及一些循环性。假设CompositeFoo的提供者如下:

public class CompositeFooProvider implements Provider<Foo> {
    @Inject private FooA first;
    @Inject @Named("Inner") private Foo second;

    @Override public Foo get() { return new CompositeFoo(asList(first, second)); }
}


提供第二个FooFooBFooC)的模块如下:

public class InnerModule extends PrivateModule {
    private final Key<? super Foo> bindingKey;  // key will be exposed, bound to the Foo below

    // configure() deals with deps of FooB and FooC

    @Provides
    public Foo getInnerFoo(...) {
        // Assume that the args are such that if they are "valid", we should return a FooB, else FooC
        if (...) return new FooB(...);
        else return new FooC(...);
    }
}


当我尝试构造外部模块时会出现这种圆形性:我需要安装InnerModule(将Key.get(Foo.class, Names.named("Inner"))作为绑定键传递)以获得第二个Foo,但是Foo已经绑定到了外部模块,因为它绑定到CompositeFooProvider。如何解决这种循环性?将@Provides方法转换为自己的Provider就足够了吗?

最佳答案

@Provides Foo方法为Foo提供绑定,该绑定与外部模块中的Foo绑定冲突。因此,将其绑定为其他内容:

public class InnerModule extends PrivateModule {
    private final Key<Foo> bindingKey;  // key will be exposed, bound to the @Inner Foo below

    @BindingAnnotation
    @Target({ FIELD, PARAMETER, METHOD })
    @Retention(RUNTIME)
    @interface Inner {
    }

    @Override
    protected void configure() {
        bind(bindingKey).to(Key.get(Foo.class, Inner.class));
        expose(bindingKey);
    }

    @Provides
    @Inner Foo getInnerFoo(...) {
        // Assume that the args are such that if they are "valid", we should return a FooB, else FooC
        if (...) return new FooB(...);
        else return new FooC(...);
    }
}


或者也许你可以做

    @Provides
    @Exposed
    @Named("Inner") Foo getInnerFoo(...) {
        // Assume that the args are such that if they are "valid", we should return a FooB, else FooC
        if (...) return new FooB(...);
        else return new FooC(...);
    }


直接并且不必费心传递绑定密钥。

10-06 13:01