当方法调用后定义隐式val时,Scala隐式参数为null

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C类中调用函数f时,该值已初始化,而在B类中,它尚未初始化. A类与B类完全相同 – 唯一的区别是Scala在A中隐式传递了arg.

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.例如,如果以B类为例,则会在调用f的位置声明val arg,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";
    }
}

使用延迟val或正确的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