在Java中实例化scala.Int

Question

我在https://github.com/play-modules/modules.playframework.org/blob/master/app/models/ss/MPOOAuth2Info.java上找到了一个例子

package models.ss;

import models.AbstractModel;
import securesocial.core.java.OAuth2Info;
import javax.persistence.Entity;

@Entity
public class MPOOAuth2Info extends AbstractModel
{
    public String accessToken;

    public String tokenType;

    public Integer expiresIn;

    public String refreshToken;

    public MPOOAuth2Info()
    {
        // no-op
    }

    public MPOOAuth2Info(OAuth2Info oAuth2Info)
    {
        this.accessToken = oAuth2Info.accessToken;
        this.tokenType = oAuth2Info.tokenType;
        this.expiresIn = oAuth2Info.expiresIn;
        this.refreshToken = oAuth2Info.refreshToken;
    }

    public OAuth2Info toOAuth2Info()
    {
        OAuth2Info oAuth2Info = new OAuth2Info();

        oAuth2Info.accessToken = this.accessToken;
        oAuth2Info.tokenType = this.tokenType;
        oAuth2Info.expiresIn = this.expiresIn;
        oAuth2Info.refreshToken = this.refreshToken;

        return oAuth2Info;
    }
}

但是接口被改变了，所以我不能使用securesocial.core.java.OAuth2Info。SecureSocial是由Scala编写的，这个类是Java的前端。所以我决定直接使用Scala：

case class OAuth2Info(accessToken: String, tokenType: Option[String] = None,
                  expiresIn: Option[Int] = None, refreshToken: Option[String] = None)

我的结果是：

package models.security.securesocial;

import models.AbstractModel;
import scala.Option;
import securesocial.core.*;

import javax.persistence.Entity;

/**
 * Persistence wrapper for SecureSocial's {@link } class.
 *
 * @author Steve Chaloner (steve@objectify.be)
 */
@Entity
public class MPOOAuth2Info extends AbstractModel
{
    public String accessToken;

    public String tokenType;

    public Integer expiresIn;

    public String refreshToken;

    public MPOOAuth2Info(){
        // no-op
    }

    public MPOOAuth2Info(OAuth2Info oAuth2Info){
        this.accessToken = oAuth2Info.accessToken();
        this.tokenType = oAuth2Info.tokenType().get();
        this.expiresIn = scala.Int.unbox(oAuth2Info.expiresIn().get());
        this.refreshToken = oAuth2Info.refreshToken().get();
    }

    public OAuth2Info toOAuth2Info(){
        return new OAuth2Info(accessToken, Option.apply(tokenType), Option.apply(SOME_TRANSFORMATION(expiresIn)), Option.apply(refreshToken));
    }
}

为了将scala.Int转换为java.lang.Integer，我使用了scala.Int.unbox()。是连接方式吗？我不知道如何将java.lang.Integer转换为scala.Int：在代码中，我输入了伪代码SOME_TRANSFORMATION()。此SOME_TRANSFORMATION的正确实现是什么？

谢谢

Answer 1

scala.Int.unbox(new java.lang.Integer(3))给了Int = 3

scala.Int.box(3)给了Integer = 3

Answer 2

Scala自动和透明地装箱和拆箱。此外，scala.Int在某种程度上是虚构的。当存储在本地变量中、作为形参传递或存储在类中时，它是原生int (类似于Java / JVM)。但是当它必须通过参数化/泛型类型的值进行传输时，必须对其进行装箱和取消装箱。(有一个例外，当有问题的类型参数是专门化的，但那是另一回事了。)