借助Play框架的JSON库,我怎么可以创建一个Reads和Writes对于没有领域一个Java枚举？
public enum EnumNoFields {
RED,
WHITE,
BLUE
}
implicit val EnumNoFieldsReads: Reads[EnumNoFields] = ?
implicit val EnumNoFieldsWrites: Writes[EnumNoFields] = ?
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？
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 …看一下Learn You a Haskell的一个kind例子:
data Frank a b = Frank {frankField :: b a} deriving (Show)
LYAH讨论:
我们看到弗兰克有一种* - >(* - >*) - >*第一个*代表a而(* - >*)代表b.
当我想到一个例子时,这对我来说很有意义:Frank String Maybe = Frank {frankField :: Maybe String}.
但我对如何思考kind右手方面感到困惑
... = Frank {frankField :: b a}.
由于返回类型frankField :: b a,怎么能kind的Frank为:
* -> (* -> *) -> * -- what does the right-most * represent?
为什么右手边kind等于* …
鉴于List[Option[Int]]:
scala> list
res8: List[Option[Int]] = List(Some(1), Some(2), None)
我可以得到List(1,2),即提取list通道flatMap和flatten:
scala> list.flatten
res9: List[Int] = List(1, 2)
scala> list.flatMap(x => x)
res10: List[Int] = List(1, 2)
鉴于[Maybe Int]Haskell中的以下内容,我该如何执行上述操作？
我尝试了下面的失败:
import Control.Monad
maybeToList :: Maybe a -> [b]
maybeToList Just x = [x]
maybeToList Nothing = []
flatten' :: [Maybe a] -> [a]
flatten' xs = xs >>= (\y -> y >>= maybeToList)
我正在使用Node-Postgres客户端连接我的 AWS Redshift 数据库。
在本地，我可以在 中运行以下代码node，获取“>>已连接”和“>>>>成功查询的打印语句。jsonResult：”。
但是，当我在 Amazon Lambda 中运行此代码时，除了“尝试连接...”之外，我没有看到任何日志语句。
console.log("trying to connect...");
var r = pg.connect(conString, function(err, client) {
if(err) {
return console.log('>> could not connect to redshift', err);
}
console.log(">> connected");
client.query('SELECT * FROM my_table', function(err, result) {
if(err) {
return console.log('error running query', err);
}
var jsonResult = JSON.stringify( result );
console.log(">>> successful query. jsonResult: " + jsonResult);
client.end();
return jsonResult;
});
});
我很困惑除了“>>尝试连接...”之外如何没有打印语句出现在这段代码中。
鉴于以下内容:
> mvn clean compile
...
[ERROR] Failure executing javac, but could not parse the error:
/bin/sh: c:/progra~1/java/jdk1.6.0_24/bin/javac: No such file or directory
检查我mvn和java版本是否正确:
$mvn -version
Apache Maven 3.3.3 (7994120775791599e205a5524ec3e0dfe41d4a06; 2015-04-22T07:57:37-04:00)
Maven home: /Users/kevin/Downloads/apache-maven-3.3.3
Java version: 1.8.0_51, vendor: Oracle Corporation
Java home: /Library/Java/JavaVirtualMachines/jdk1.8.0_51.jdk/Contents/Home/jre
Default locale: en_US, platform encoding: UTF-8
OS name: "mac os x", version: "10.10.4", arch: "x86_64", family: "mac"
$java -version
java version "1.8.0_51"
Java(TM) SE Runtime Environment (build 1.8.0_51-b16)
Java HotSpot(TM) 64-Bit Server VM …鉴于以下特征:
scala> trait Foo { self =>
| def f: String = self.getClass.getName
| }
defined trait Foo
scala> trait Bar {
| def f: String = this.getClass.getName
| }
defined trait Bar
然后创建扩展它们的类:
scala> class FooImpl extends Foo {}
defined class FooImpl
scala> class BarImpl extends Bar {}
defined class BarImpl
然后f在新实例上调用它们的方法:
scala> (new FooImpl).f
res1: String = FooImpl
scala> (new BarImpl).f
res4: String = BarImpl
REPL显示它们打印出相同的值 - 类的名称.
也许这不是一个好例子.但是,什么是使用的差异self在上面Foo相比Bar,它使用this …
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？