当方法调用后定义隐式val时,Scala隐式参数为null

Question

考虑:

object implicitnull extends App {
  mymethod

  implicit val arg = "foo"

  def mymethod(implicit arg: String) = {
    arg.size
  }
}

但是,这不会导致任何编译错误,而是在运行时导致NullPointerException来自arg.size.

这是预期的行为吗？

Answer 1

是的,这是由于Scala构造类并初始化它们的方式所预期的行为.考虑这个例子:

scala> class A {
     |   f
     | 
     |   implicit val arg = "foo"
     | 
     |   def f(implicit arg: String) = {
     |     println(arg)
     |   }
     | }
defined class A

scala> class B {
     |   f(arg)
     | 
     |   val arg = "foo"
     | 
     |   def f(implicit arg: String) = {
     |     println(arg)
     |   }
     | }
defined class B

scala> class C {
     |   implicit val arg = "foo"
     | 
     |   f
     | 
     |   def f(implicit arg: String) = {
     |     println(arg)
     |   }
     | }
defined class C

scala> new A
null
res0: A = A@67d3caf

scala> new B
null
res1: B = B@3f2c5ad4

scala> new C
foo
res2: C = C@177bdd23

在f类中调用该函数时,C该值已初始化,而在类中B它尚未初始化.类A与类完全相同B- 唯一的区别是Scala arg隐式传递A.

这有点让人困惑,因为这段代码正在做两件事 - 它声明成员变量并执行构造函数代码.B例如,如果以类为例,val arg将在f被调用的位置声明,但尚未初始化.val arg = "foo"进行初始化.一旦你将它翻译成Java,它就会变得更加明显:

public class B {

    void f(String arg) {
        System.out.println(arg);
    }

    String arg; // also acts as final in Scala

    public B() {
        f(arg);
        arg = "foo";
    }
}

使用lazy val或更正初始化顺序将有助于解决它:

scala> class B {
     |   f(arg)
     | 
     |   lazy val arg = "foo"
     | 
     |   def f(implicit arg: String) = {
     |     println(arg)
     |   }
     | }
defined class B

scala> new B
foo
res3: B = B@3f9ac6e6