类不是抽象的，不实现抽象基类成员

Question

我对这个与为从 Maven 包导入的抽象类提供实现相关的 Kotlin 错误感到困惑。

我有一个用 Kotlin 编写的 Maven 库，并公开了一个名为 APIGatewayRequestHandler 的抽象类。在导入库的应用程序中，我提供了抽象类的实现：

class GetWelcomeMessageHandler : APIGatewayRequestHandler<WelcomeMessage>()
    fun handleAPIGatewayRequest(input: com.amazonaws.services.lambda.runtime.events.APIGatewayProxyRequestEvent, context: com.amazonaws.services.lambda.runtime.Context?): WelcomeMessage {
        return WelcomeMessage()
    }
}

从库中反编译的抽象类如下所示：

public abstract class APIGatewayRequestHandler<T> public constructor() : com.amazonaws.services.lambda.runtime.RequestHandler<com.amazonaws.services.lambda.runtime.events.APIGatewayProxyRequestEvent, T> {
    public abstract fun handleAPIGatewayRequest(input: com.amazonaws.services.lambda.runtime.events.APIGatewayProxyRequestEvent, context: com.amazonaws.services.lambda.runtime.Context?): T

    public open fun handleRequest(input: com.amazonaws.services.lambda.runtime.events.APIGatewayProxyRequestEvent?, context: com.amazonaws.services.lambda.runtime.Context?): T {
        /* compiled code */
    }
}

我收到以下错误：

Class 'GetWelcomeMessageHandler' is not abstract and does not implement abstract base class member
public abstract fun handleAPIGatewayRequest(input: APIGatewayProxyRequestEvent, context: Context?): WelcomeMessage

Answer 1

我认为您只是缺少一些override关键字。也就是说，你的抽象类应该在handleRequest方法上有它：

public abstract class APIGatewayRequestHandler<T> public constructor() : com.amazonaws.services.lambda.runtime.RequestHandler<com.amazonaws.services.lambda.runtime.events.APIGatewayProxyRequestEvent, T> {
    public abstract fun handleAPIGatewayRequest(input: com.amazonaws.services.lambda.runtime.events.APIGatewayProxyRequestEvent, context: com.amazonaws.services.lambda.runtime.Context?): T

    public override fun handleRequest(input: com.amazonaws.services.lambda.runtime.events.APIGatewayProxyRequestEvent?, context: com.amazonaws.services.lambda.runtime.Context?): T {
        /* compiled code */
    }
}

然后你GetWelcomeMessageHandler应该有它的handleAPIGatewayRequest方法：

class GetWelcomeMessageHandler : APIGatewayRequestHandler<WelcomeMessage>() { // <-- This curly brace was also missing
    override fun handleAPIGatewayRequest(input: com.amazonaws.services.lambda.runtime.events.APIGatewayProxyRequestEvent, context: com.amazonaws.services.lambda.runtime.Context?): WelcomeMessage {
        return WelcomeMessage()
    }
}