object Test {
def main(args: Array[String]) {
val list: List[Double] = List(1.0, 2.0, 3.0, 4.0)
val none = None
case class Test()
val test = Test()
def f(x: Any) = x match {
case _: Some[Test] => println("_ matched")
case None => println("None matched")
}
f(list)
f(none)
f(test)
}
}
尝试编译上面的代码会导致"擦除"编译时警告.
$>scalac Test.scala
Test.scala:11: warning: non-variable type argument Test in type pattern
Some[Test] is unchecked since it is eliminated by erasure
case _: Some[Test] => println("_ matched")
^
one warning …我打电话tail给Array,但看到了一个警告.
scala> val arr = Array(1,2)
arr: Array[Int] = Array(1, 2)
scala> arr tail
warning: there were 1 feature warning(s); re-run with -feature for details
res3: Array[Int] = Array(2)
Scaladocs for Array显示了一个UnsupportedOperationException [will be thrown]
if the mutable indexed sequence is empty.
是否存在安全,不会抛出异常,获取数组尾部的方法？
在另一个练习中实现Monad.sequence()从斯卡拉函数式编程,我的回答来自官方不同/知到是正确的答案:
def序列[A](lma:List [F [A]]):F [List [A]]
官方:
def sequence[A](lma: List[F[A]]): F[List[A]] =
lma.foldRight(unit(List[A]()))((ma, mla) => map2(ma, mla)(_ :: _))
矿:
def sequence[A](lma: List[F[A]]): F[List[A]] = F(lma.flatten)
F的例子是Option:
scala> val x: List[Option[Int]] = List( Some(1), None)
x: List[Option[Int]] = List(Some(1), None)
scala> Some(x.flatten)
res1: Some[List[Int]] = Some(List(1))
我的回答(或其精神)在这里合法吗？
我得到以下编译时异常,但我确定它是否与我对类型构造函数缺乏理解有关.
Monad.scala:15:错误:未找到:值F
F(lma.flatten)
阅读Effective Java，我从Item 16: Favor composition over inheritance.
在下面InstrumentedSet，这本书显示我们可以跟踪元素被插入的次数（通过InstrumentedSet.addCount变量）。
要做到这一点，我们可以简单地附加到这个类对象的addCount，然后调用ForwardingSet.add()，它调用实际Set类的 实际实现add()。
// Reusable forwarding class
public class ForwardingSet<E> implements Set<E> {
private final Set<E> s;
public ForwardingSet(Set<E> s) { this.s = s; }
public void clear() { s.clear(); }
public boolean contains(Object o) { return s.contains(o); }
...
}
// Wrapper class - uses composition in place of inheritance
public class InstrumentedSet<E> extends ForwardingSet<E> { …我Named Arguments在Scala中深入研究这个例子:
scala> class Parent {
| def foo(bar: Int = 1, baz: Int = 2): Int = bar + baz
| }
defined class Parent
scala> class Child extends Parent {
| override def foo(baz: Int = 3, bar: Int = 4): Int = super.foo(baz, bar)
| }
defined class Child
scala> val p = new Parent
p: Parent = Parent@6100756c
scala> p.foo()
res1: Int = 3
scala> val x = new Child
x: Child …请考虑ui-router的wiki中的以下略微修改的代码.
var myApp = angular.module('myApp',['ui.router']);
myApp.config(function($stateProvider, $urlRouterProvider, $locationProvider) {
// for any unmatched url
$urlRouterProvider.otherwise("/state1");
$locationProvider.html5Mode(true);
// now set up the states
$stateProvider
.state('state1', {
url: '/state1',
templateUrl: "partials/state1.html"
})
.state('state1.list', {
url: "/list",
templateUrl: "partials/state1.list.html",
controller: function($scope) {
$scope.items = ["A", "List", "of", "Items"];
}
})
.state('state2', {
url: "/state2",
templateUrl: "partials/state2.list.html",
controller: function($scope) {
$scope.things = ["a", "set", "of", "things"];
}
})
});
运行python 3的SimpleHttpServer,我404 error在尝试访问时得到:http://localhost:8000/state1234324324.
为什么没有$urlRouterProvider.otherwise("/state1");重定向所有未知路由/state1,根据这个问题,已经定义state …
给定以下代数数据类型：
scala> sealed trait Person
defined trait Person
scala> case class Boy(name: String, age: Int, x: String) extends Person
defined class Boy
scala> case class Girl(name: String, age: Int, y: Boolean) extends Person
defined class Girl
注意 - 我知道它不是递归类型 - 不涉及递归。
那么，这是 aSum还是Product Type？为什么？
给出以下AST Success和Failure:
sealed trait Success
case object FooGood extends Success
case object BarGood extends Success
sealed trait Failure
case object FooBad extends Failure
case object BarBad extends Failure
方法签名:
def go[A <: Failure, B <: Success](x: Int): Either[A, B] = ???
但是,我想约束Left和Right类型特定于Foo或Bar.
但是下面的代码编译(违背我的意愿):
scala> go[FooBad.type, BarGood.type](5)
scala.NotImplementedError: an implementation is missing
如何在编译时实现此约束？
在Idris中,我可以通过以下方式添加两个相同大小的向量:
module MatrixMath
import Data.Vect
addHelper : (Num n) => Vect k n -> Vect k n -> Vect k n
addHelper = zipWith (+)
在REPL上编译后:
*MatrixMath> :l MatrixMath.idr
Type checking ./MatrixMath.idr
然后我可以用两个大小为3的向量来调用它:
*MatrixMath> addHelper [1,2,3] [4,5,6]
[5, 7, 9] : Vect 3 Integer
但是,当我尝试调用addHelper两个不同大小的向量时,它将无法编译:
*MatrixMath> addHelper [1,2,3] [1]
(input):1:20:When checking argument xs to constructor Data.Vect.:::
Type mismatch between
Vect 0 a (Type of [])
and
Vect 2 n (Expected type)
Specifically:
Type mismatch between
0
and
2 …使用Idris进行类型驱动开发提供了以下通用加法器方法:
AdderType : (numArgs : Nat) -> Type
AdderType Z = Int
AdderType (S k) = (next : Int) -> AdderType k
adder : (n : Nat) -> (acc : Int) -> AdderType n
adder Z acc = acc
adder (S k) acc = \x => (adder k (x+acc))
例:
-- expects 3 Int's to add, with a starting value of 0
*Work> :t (adder 3 0)
adder 3 0 : Int -> Int -> Int -> Int
-- …