看看ClassTag#runtimeClass,它有一个返回类型Class[_],即Class根据我的理解,带有一个通配符参数.
我试图实现一个方法A => ClassTag[A]:
import scala.reflect._
scala> def f[A](x: A)(implicit ev: ClassTag[A]) = ev.runtimeClass
f: [A](x: A)(implicit ev: scala.reflect.ClassTag[A])Class[_]
但是,def正如文档所示,该定义的输出是Class[_].
是否有可能改变f它的返回类型Class[A]？如果没有,那为什么不可能呢？
鉴于:
*Main> let s = state $ \x -> ("foo", x)
*Main> :t s
s :: MonadState s m => m [Char]
我试着打电话get s,但得到了以下编译时错误:
*Main> :t get
get :: MonadState s m => m s
*Main> let x = get s
<interactive>:95:5: error:
• Non type-variable argument
in the constraint: MonadState t ((->) (m [Char]))
(Use FlexibleContexts to permit this)
• When checking the inferred type
x :: forall s (m :: * -> *) t.
(MonadState s …我正在尝试编写一个函数,StateT只是为了更多地了解它.
在f,我想访问Int最后一个类型的参数StateT [Int] IO Int:
f :: StateT [Int] IO Int
f = state $ \xs -> update (error "I want a") xs
update :: Int -> [Int] -> (Int, [Int])
update x [] = (x, [])
update x (y:ys) = (x+y, ys)
以下是我想称之为:
let x = return 55 :: StateT [Int] IO Int
参考runStateT:
*Main> :t runStateT
runStateT :: StateT s m a -> s -> m (a, s) …鉴于以下内容:
$cat src/Greeting.re
let component = ReasonReact.reducerComponent("Greeting");
type action =
| Click;
type state = {
count: int
};
let make = (_children) => {
...component,
initialState: () => {count: 0},
reducer: (action, state) =>
ReasonReact.Update({count: state.count + 1}),
render: (self) => {
let message = "Clicked " ++ string_of_int(self.state.count) ++ "x";
<div>
<button
onClick={_event => self.send(Click)}
/>
{ReasonReact.stringToElement(message)}
</div>
}
};
我得到以下编译时错误:
17 ? <div>
18 ? <button
19 ? onClick={_event => self.send(Click)}
20 ? />
21 ? {ReasonReact.stringToElement(message)} …我如何能明确标记的类型let myList = [1, 2, 3];为list of int's？
我let xs: 'int list = [1,2,3];通过Try ReasonML尝试失败了.
为什么以下两个foldLeft产生相同的输出？
#1
scala> List(1,2,3).foldLeft(List[Int]())( (acc, el) => acc :+ el)
res114: List[Int] = List(1, 2, 3)
并且,现在使用_ :+ _作为(B, A) => B参数.
#2
scala> List(1,2,3).foldLeft(List[Int]())(_ :+ _)
res115: List[Int] = List(1, 2, 3)
特别是,accumulator在第二种情况下缺乏明确附加的内容使我感到困惑.
我正在尝试过滤list of 2-tuples第一个元组值等于的位置0:
ghci> ys
[(0,55),(1,100)]
ghci> filter (\x -> x.fst == 0) ys
<interactive>:71:27:
Couldn't match type `(Integer, Integer)' with `b0 -> c0'
Expected type: [b0 -> c0]
Actual type: [(Integer, Integer)]
In the second argument of `filter', namely `ys'
In the expression: filter (\ x -> x . fst == 0) ys
In an equation for `it': it = filter (\ x -> x . fst == 0) ys
我想要的输出:
[(1,100)]
我怎样才能做到这一点？另外,编译时错误是什么意思？
鉴于:
scala> case class Foo(x: Int, y: String)
defined class Foo
我正在尝试Foo.tupled用来创建Function2[Int, String, Foo]:
scala> val fn2: Function2[Int, String, Foo] = Foo.tupled match {
| case (param1, param2) => { (param1, param2) => Foo(param1, param2) }
| }
<console>:18: error: constructor cannot be instantiated to expected type;
found : (T1, T2)
required: ((Int, String)) => Foo
case (param1, param2) => { (param1, param2) => Foo(param1, param2) }
但是,它不起作用.我该如何解决这个破碎的代码？
我可以通过使用申请获得该index = 2项目.List
scala> List(1,2,3).apply(2)
res3: Int = 3
scala> val x = List(1,2,3)
x: List[Int] = List(1, 2, 3)
scala> x(2)
res4: Int = 3
scala> List(1,2,3).apply(2)
res5: Int = 3
但是,为什么我不能做以下事情呢？
scala> List(1,2,3).(2)
<console>:1: error: identifier expected but '(' found.
List(1,2,3).(2)
^
给出以下方法:
scala> def f: List[Any] => Any = xs => 1234 // this output does not matter
f: List[Any] => Any
模式匹配是否可能List[Any] => Any？我没有unapply在Function1上看到一个方法,所以我认为答案是否定的.
这是我正在尝试做的事情:
例:
def foo(x: Any) = x match {
case ... // to handle the case of List[Any] => Any]?
case ...
}
也许我可以计算出arity的x: Any区分List[Any] => Any与其他一切(_)？
编辑:
我希望我不必依赖f.toString== <function1>.