scala Any 类型的表达式不符合预期的 type_$1
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/37860513/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Expression of type Any doesn't conform to expected type_$1
提问by Daryl
I have problems with undestanding Scala's type system.
我无法理解 Scala 的类型系统。
class A[T](var value: T)
class B[T](val a: A[T], var newValue: T)
val list = new MutableList[B[_]]()
val a: A[Int] = new A(10)
val b: B[Int] = new B(a, 11)
list += b
val someB = list.head
someB.a.value = someB.newValue
And after compile i see the error:
编译后我看到错误：
Error:(12, 24) type mismatch;
found : A$A36.this.someB.newValue.type (with underlying type _)
required: _
someB.a.value = someB.newValue
^
Both someB.a.valueand someB.newValuehave the same type but Scala's compiler actually doesn't think so. How this error can be fixed ?
双方someB.a.value并someB.newValue具有相同的类型，但Scala的编译器实际上并不这么认为。如何修复此错误？
采纳答案by Alexey Romanov
Don't expect the compiler to figure out two existentials are the same, even if they clearly have to be. Workarounds:
不要指望编译器会发现两个存在项是相同的，即使它们显然必须是相同的。解决方法：
Use a type variable pattern:
list.head match { case someB: B[a] => someB.a.value = someB.newValue }Extract a method:
def setValue[A](b: B[A]) = b.a.value = b.newValue setValue(someB)
使用类型变量模式：
list.head match { case someB: B[a] => someB.a.value = someB.newValue }提取方法：
def setValue[A](b: B[A]) = b.a.value = b.newValue setValue(someB)
回答by Daryl
One of solutions I have found is to eliminate usage of existential type with abstract types.
我发现的解决方案之一是消除对抽象类型的存在类型的使用。
class A[T](var value: T)
abstract class B {
type T
val a: A[T]
val newValue: T
}
val list = new mutable.MutableList[B]()
val aa = new A(10)
val b = new B {
type T = Int
val a = aa
val newValue = 11
}
list += b
val someB = list.head
someB.a.value = someB.newValue
