在Scala中，是否可以实例化通用类型T的对象? [英] In Scala, is it possible to instantiate an object of generic type T?

问题描述

在Scala中，即使解决方案不是很好，也可以实例化/创建泛型T的新对象吗?可以使用反射来实现吗?

In Scala, even if the solution is not elegant, is it possible to instantiate/create a new object of a generic type T? Is it possible to achieve this using reflection?

例如，我对以下内容感兴趣:

For example, I am interested in something like the following:

    case class Person(name: String, age: Int)

假设我要执行以下操作来创建类型为Person的对象:

Let's say I wanted to do the following to create an object of type Person:

    def createObject[T](fieldValues: Seq[Any]): T = {
        ... T(fieldValues)
    }

    val person = createObject[Person](Seq("Bob", 20))

推荐答案

从技术上讲，可以使用反射.例如，您可以使用ClassTag捕获类型为T的运行时类，然后找到合适的构造函数并创建实例:

Technically speaking it's possible using reflection. You could for example catch runtime class of type T using ClassTag then find proper constructor and create instance:

def createObject[T](fieldValues: Seq[Any])(implicit ct: ClassTag[T]): Option[T] = {
   //we lookup for matching constructor using arguments count, you might also consider checking types
   ct.runtimeClass.getConstructors.find(_.getParameterCount == fieldValues.size) 
   .flatMap {constructor =>
        Try(constructor.newInstance(fieldValues: _*).asInstanceOf[T]).toOption
    }
}

createObject[Person](Seq("Bob", 20)) //Some(Person("Bob", 20))
createObject[Person](Seq(20, 10)) //None

如果没有构造函数匹配的参数，则该函数将无法返回None.

In case there's no constructor matching parameters, that function fails returning None.

它应该可以工作，但是最好避免这种方法，因为这样会失去所有类型的安全性.

It should work, but it'd be best if you can avoid this approach because you lose all type safety.

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