同一类Kotlin中的@Provides和@Binds方法

在dagger 2.11之后,我们可以使用@Binds注释并在这种情况下将我们的模块标记为抽象,这比具体更有效.

如果我的模块同时具有@Provides和@Binds方法,我有两个选择:

最简单的方法是将@Provides实例方法标记为静态.

如果有必要将它们保留为实例方法,则可以将模块拆分为两个,并将所有@Binds方法提取到抽象模块中.

第二个选项在Java和Kotlin中运行良好,但第一个选项在Java中工作正常,但我不知道如何在Kotlin中实现相同的功能.如果我将@Provides方法移动到Companion对象,它会抛出Error:(30, 1) error: @Provides methods can only be present within a @Module or @ProducerModule.怎么能在Kotlin做到这一点.

第二种选择:(工作)

ApplicationModule.kt

@Module(includes = [ApplicationModule.Declarations::class])
abstract class ApplicationModule {
    @Module
    internal interface Declarations {
        @Binds
        fun bindContext(application: Application): Context
    }

    @Provides
    @Singleton
    fun provideMvpStarterService(): MvpStarterService {
        return MvpStarterServiceFactory.makeStarterService()
    }
}

第一种选择:(不工作)

ApplicationModule.kt

@Module
abstract class ApplicationModule {
    //expose Application as an injectable context
    @Binds
    internal abstract fun bindContext(application: Application): Context

    companion object {
        @JvmStatic
        @Provides
        @Singleton
        fun provideMvpStarterService(): MvpStarterService {
            return MvpStarterServiceFactory.makeStarterService()
        }
    }
}

为第一个选项生成的Java文件:

@kotlin.Metadata(mv = {1, 1, 9}, bv = {1, 0, 2}, k = 1, d1 = {"\u0000\u001a\n\u0002\u0018\u0002\n\u0002\u0010\u0000\n\u0002\b\u0002\n\u0002\u0018\u0002\n\u0000\n\u0002\u0018\u0002\n\u0002\b\u0003\b\'\u0018\u0000 \b2\u00020\u0001:\u0001\bB\u0005\u00a2\u0006\u0002\u0010\u0002J\u0015\u0010\u0003\u001a\u00020\u00042\u0006\u0010\u0005\u001a\u00020\u0006H!\u00a2\u0006\u0002\b\u0007\u00a8\u0006\t"}, d2 = {"Lio/mywebsie/di/ApplicationModule;", "", "()V", "bindContext", "Landroid/content/Context;", "application", "Landroid/app/Application;", "bindContext$app_debug", "Companion", "app_debug"})
@dagger.Module()
public abstract class ApplicationModule {
    public static final io.mywebsie.di.ApplicationModule.Companion Companion = null;

    @org.jetbrains.annotations.NotNull()
    @dagger.Binds()
    public abstract android.content.Context bindContext$app_debug(@org.jetbrains.annotations.NotNull()
    android.app.Application application);

    public ApplicationModule() {
        super();
    }

    @org.jetbrains.annotations.NotNull()
    @javax.inject.Singleton()
    @dagger.Provides()
    public static final io.mywebsie.data.remote.MvpStarterService provideMvpStarterService() {
        return null;
    }

    @kotlin.Metadata(mv = {1, 1, 9}, bv = {1, 0, 2}, k = 1, d1 = {"\u0000\u0012\n\u0002\u0018\u0002\n\u0002\u0010\u0000\n\u0002\b\u0002\n\u0002\u0018\u0002\n\u0000\b\u0086\u0003\u0018\u00002\u00020\u0001B\u0007\b\u0002\u00a2\u0006\u0002\u0010\u0002J\b\u0010\u0003\u001a\u00020\u0004H\u0007\u00a8\u0006\u0005"}, d2 = {"Lio/mywebsie/di/ApplicationModule$Companion;", "", "()V", "provideMvpStarterService", "Lio/mywebsie/data/remote/MvpStarterService;", "app_debug"})
    public static final class Companion {

        @org.jetbrains.annotations.NotNull()
        @javax.inject.Singleton()
        @dagger.Provides()
        public final io.mywebsie.data.remote.MvpStarterService provideMvpStarterService() {
            return null;
        }

        private Companion() {
            super();
        }
    }
}

更新:

感谢@David Medenjak,您在评论中提供的链接使一切都清楚我遇到了两种方法来实现第一个选项.

更新的代码:

第一选择:(工作)

ApplicationModule.kt

@Module(includes = [ApplicationModule.AModule::class])
abstract class ApplicationModule {

    @Binds
    abstract fun bindContext(application: Application): Context

    @Module
    object AModule {
        @JvmStatic
        @Provides
        @Singleton
        fun provideMvpStarterService(): MvpStarterService {
            return MvpStarterServiceFactory.makeStarterService()
        }
    }
}

要么

@Module
abstract class ApplicationModule {
    @Binds
    abstract fun bindContext(application: Application): Context

    @Module
    companion object {
        @Provides
        @Singleton
        fun provideMvpStarterService(): MvpStarterService {
            return MvpStarterServiceFactory.makeStarterService()
        }
    }
}

两种工作都很好,但由于某种原因,第一个选项看起来不吸引我,所以我更喜欢第二个选项.