从Scala中的Functional Programming,我正在尝试实现Either.map.
trait Either[+E, +A] {
def map[B](f: A => B): Either[E, B] = this match {
case Either(x: E, y: A) => Right(f(y))
case _ => Left()
}
}
编译中出现一个错误,其中包括.我没有展示它们,因为我似乎错过了实施的概念Either.
Either.scala:3: error: value Either is not a case class constructor,
nor does it have an unapply/unapplySeq method
case Either(x: E, y: A) => Right(f(y))
请告知我实施它.
深度Scala在mutability和上呈现此代码equality.
class Point2(var x: Int, var y: Int) extends Equals {
def move(mx: Int, my: Int) : Unit = {
x = x + mx
y = y + my
}
override def hashCode(): Int = y + (31*x)
def canEqual(that: Any): Boolean = that match {
case p: Point2 => true
case _ => false
}
override def equals(that: Any): Boolean = {
def strictEquals(other: Point2) =
this.x == other.x && this.y == other.y
that …为什么不window.x打印出来10？
eval("var x = 10;");
console.log(window.x); // undefined
console.log(x); // 10
使用IntelliJ的12 Ultimate,我在调试器中运行以下代码:
Java的
import play.api.libs.json.JsValue;
public class Foo {
...
public JsValue toJson() { ... }
public class FooExample {
...
Foo foo = new Foo();
System.out.println("...); //<-- breakpoint
在断点处,我右键单击我的源代码并选择"评估表达式",输入:
foo.toJson().
但出现以下错误:
No such instance method: play.api.libs.json.JsValue$class.com.foo.Foo.toJson ()
难道我做错了什么？Foo#toJson调用Scala代码,如果重要的话.
编辑我实际上在实例化之后有了断点.Foo.对于那些投票的人,我当之无愧.
查看Scala中的Functional Programming中的IO Monad示例:
def ReadLine: IO[String] = IO { readLine }
def PrintLine(msg: String): IO[Unit] = IO { println(msg) }
def converter: IO[Unit] = for {
_ <- PrintLine("enter a temperature in degrees fahrenheit")
d <- ReadLine.map(_.toDouble)
_ <- PrintLine((d + 32).toString)
} yield ()
我决定converter用一个重写flatMap.
def converterFlatMap: IO[Unit] = PrintLine("enter a temperate in degrees F").
flatMap(x => ReadLine.map(_.toDouble)).
flatMap(y => PrintLine((y + 32).toString))
当我用最后一个替换flatMap时map,我没有看到在控制台上打印出readLine的结果.
用flatMap:
enter a …考虑以下Java ArrayList#toArray方法测试.请注意,我从这个有用的答案中借用了代码.
public class GenericTest {
public static void main(String [] args) {
ArrayList<Integer> foo = new ArrayList<Integer>();
foo.add(1);
foo.add(2);
foo.add(3);
foo.add(4);
foo.add(5);
Integer[] bar = foo.toArray(new Integer[10]);
System.out.println("bar.length: " + bar.length);
for(Integer b : bar) { System.out.println(b); }
String[] baz = foo.toArray(new String[10]); // ArrayStoreException
System.out.println("baz.length: " + baz.length);
}
}
但是,请注意,ArrayStoreException在尝试Integer进入时会有一个时间String[].
输出:
$>javac GenericTest.java && java -cp . GenericTest
bar.length: 10
1
2
3
4
5
null
null
null …我想创建kevin一个密码为的新用户kevin.
食谱/ make_user /食谱/ default.rb
user "kevin" do
comment "default user"
home "/home/kevin"
shell "/bin/bash"
password "kevin"
end
在配置我的Vagrant盒子之后,我将其放入盒子中.然而,我无法su为kevin与密码的用户kevin.
[vagrant@vagrant-centos65 ~]$ su kevin
Password:
su: incorrect password
查看Chef'用户' 文档,我不确定这password是否是要修改的正确属性.
password The password shadow hash. This attribute requires that
ruby-shadow be installed. This is part of the Debian package:
libshadow-ruby1.8.
如何修改我上面的食谱,这样我可以su为kevin使用相同的密码？
break[a] -> (a -> Bool) -> ([a], [a])根据我的理解,有第一个元组等于的签名takeWhile predicate is true.第二元组是负责使谓词为假加上剩余列表的项目.
> break (== ' ') "hey there bro"
("hey"," there bro")
但是,是否有一个功能会跳过负责破坏的项目？
>foo? (== ' ') "hey there bro"
("hey","there bro")
toUpper c = chr (fromIntegral (towupper (fromIntegral (ord c))))
...
foreign import ccall unsafe "u_towupper"
towupper :: CInt -> CInt
是什么意思chr,以及u_towupper？我对这foreign import ccall unsafe部分也很好奇.Haskell源实际上是否会发生变异,因此unsafe？
我正在尝试使用Object.defineProperty更新我的obj对象以使用get和set访问器obj.name.
var obj = {};
Object.defineProperty(obj, 'name', {
get: function() { return this.name; },
set: function(x) { this.name = x; }
});
console.log("obj:", obj);
console.log("obj.name:", obj.name);
但是我得到了一个Uncaught RangeError: Maximum call stack size exceeded.
如何使用在属性中Object.defineProperty添加get和set访问者？nameobj