[英]Scala 'type mismatch error' while using a function but works fine for method
def stringToIntMethod(input:String):Option[Int] = {
try{
Some(Integer.parseInt(input.trim()))
}
catch{
case e:Exception => None
}
}
val stringToIntFunction: (String) => Option[Int] = (in:String) => {
try{
Some(Integer.parseInt(in.trim()))
}
catch{
case e:Exception => None
}
}
val stringAndIntArray = Array("Hello", "1","2","Hi") //Input
println("with Method is: " + stringAndIntArray.flatMap(stringToIntMethod))
println("with functon is: " + stringAndIntArray.flatMap(stringToIntFunction))
getting a type mismatch error while using stringToIntFunction in flatMap 在flatMap中使用stringToIntFunction时遇到类型不匹配错误
type mismatch;
found : String => Option[Int]
required: String => scala.collection.GenTraversableOnce[?]
println("with functon is: " + stringAndIntArray.flatMap(stringToIntFunction))
^
Why is it? 为什么?
flatMap
requires lambda, you are passing it a normal method flatMap
需要lambda,您将其传递给普通方法
Here is the fix 这是解决方法
stringAndIntArray.flatMap(stringToIntMethod _)
scala> def toInt(s: String): Option[Int] = Some(s.toInt)
toInt: (s: String)Option[Int]
scala> Array("1", "2", "3").flatMap(toInt _)
res1: Array[Int] = Array(1, 2, 3)
Normal method can be converted to lambda using underscore 普通方法可以使用下划线转换为lambda
scala> def foo(s: String, i: Int): Double = 1
foo: (s: String, i: Int)Double
scala> foo _
res2: (String, Int) => Double = $$Lambda$1162/1477996447@62faf77
scala> foo(_, _)
res3: (String, Int) => Double = $$Lambda$1168/1373527802@30228de7
scala> foo(_: String, _: Int)
res5: (String, Int) => Double = $$Lambda$1183/477662472@2adc1e84
scala> foo("cow", _: Int)
res7: Int => Double = $$Lambda$1186/612641678@146add7b
scala> foo("Cow is holy", _: Int)
res8: Int => Double = $$Lambda$1187/1339440195@7d483ebe
Also adding a comment from lambda. 还添加了来自lambda的评论。 xy.
XY。 x
X
f _ is syntactic sugar for f(_) which is again syntactic sugar for x => f(x)
I'm still puzzled myself about why this example does not work. 我仍然对为什么这个例子不起作用感到困惑。 I assume it has to do with Scala's type inference.
我认为这与Scala的类型推断有关。 If you look at the error message
如果您查看错误消息
found : String => Option[Int]
required: String => scala.collection.GenTraversableOnce[?]
it says that the return value of stringToIntFunction does not fit the argument value of flatMap
. 它说stringToIntFunction的返回值不适合
flatMap
的参数值。 Indeed, the type test 确实,类型测试
Some(1).isInstanceOf[TraversableOnce[Int]]
leads to: 导致:
<console>:138: warning: fruitless type test: a value of type Some[Int] cannot also be a scala.collection.TraversableOnce[Int] (the underlying of TraversableOnce[Int])
Some(1).isInstanceOf[TraversableOnce[Int]]
^
res24: Boolean = false
Funnily enough, when I change the return type of your function to TraversableOnce[Int]
, it works: 有趣的是,当我将函数的返回类型更改为
TraversableOnce[Int]
,它会起作用:
def stringToIntFunction : String => TraversableOnce[Int] = (in:String) => {
//...
}
leads to 导致
scala> stringAndIntArray.flatMap(stringToIntFunction)
res28: Array[Int] = Array(1, 2)
The reason is that even though Option
does not derive from TraversableOnce
, there is an implicit conversion: 原因是,即使
Option
不是从TraversableOnce
派生的,也存在隐式转换:
scala> def f(to : TraversableOnce[Int]) = to.size
f: (to: TraversableOnce[Int])Int
scala> f(Some(1))
res25: Int = 1
This has also been noticed in a different question before. 之前在另一个问题中也注意到了这一点。
My theory is that for a method the return value is explicitly known to the compiler which allows it to detect the presence of an implicit conversion from Option[Int] => TraversableOnce[Int]
. 我的理论是,对于一个方法,返回值对于编译器来说是明确已知的,这使它可以检测到
Option[Int] => TraversableOnce[Int]
隐式转换的存在。 But in the case of a function value the compiler would only look for an implicit conversion between (String => Option[Int]) => (String => TraversableOnce[Int])
. 但是对于函数值,编译器只会在
(String => Option[Int]) => (String => TraversableOnce[Int])
之间寻找隐式转换。 When the lambda application stringToIntFunction _
is passed instead, the compiler seems to see that it can apply the implicit conversion again. 相反,当传递lambda应用程序
stringToIntFunction _
,编译器似乎看到它可以再次应用隐式转换。
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