I have the following list defined within a Scala object:

object Foo {
    val bar = List(1, 2, 3)
}

Which seems to become a scala.collection.immutable.List&l,java,scala"/>

时间戳

首页 > 解决方案 > Scala List[Int] 变为 List in Java

I have the following list defined within a Scala object:

object Foo {
    val bar = List(1, 2, 3)
}

Which seems to become a scala.collection.immutable.List&l

java - Scala List[Int] 变为 List in Java

I have the following list defined within a Scala object:

object Foo {
    val bar = List(1, 2, 3)
}

Which seems to become a scala.collection.immutable.List&l

问题描述

I have the following list defined within a Scala object:

object Foo {
    val bar = List(1, 2, 3)
}

Which seems to become a scala.collection.immutable.List<Object> when I try to use it from Java. As a consequence, I have to use getters like Foo.bar().apply$mcII$sp(i) or apply with a cast to Integer/int.

Why is the generic type Object and not Integer? This also seems to be only the case for types that exist as Java primitives; List[MyType] becomes List<MyType> in Scala.

Append report parameter in ssrs

I have two parameters in the SSRS report(Company and Department).'Department' Parameter is dependent on the value you select from Company. For Example, Company :A --> Department:x or Department:y (Company A has 2 department x and y)

I am trying to append the parameters in the Report URL by using &rp:Company=A this works fine.

But when i try to append the Second Parameter &rp:Company=A&Department=x in the URL it does not pick the second parameter value ,i.e. it selects Company as 'A' but unable to select the Department as 'x'.

what would be a correct way to append second parameter?

I have tried using &rs:Command=Render&Company=A&Department=x but this does not work.

标签： javascala

解决方案

发生这种情况是因为Java 不支持泛型类型中的原始类型，泛型仅在 Java 中是编译时构造，它们不存在于 JVM 字节码中，因此它们必须可转换为 Java 的Object类型，而原始类型不能。

如果我们使用-Xprint:jvm标志编译代码，您可以看到它List[Int]实际上编译为非泛型List：

package com.yuvalitzchakov.github {
  object Foo extends Object {
    private[this] val bar: List = _;
    <stable> <accessor> def bar(): List = Foo.this.bar;
    def <init>(): com.yuvalitzchakov.github.Foo.type = {
      Foo.super.<init>();
      Foo.this.bar = scala.collection.immutable.List.apply(scala.Predef.wrapIntArray(Array[Int]{1, 2, 3}));
      ()
    }
  }
}

如果Foo.bar是 a List[Integer]，这将List<Integer>在 Java 中产生 a，而不是List<Object>

推荐阅读