traits

问题 I am frequently associating a vote entity in other ones with symfony2 / doctrine 2. This is done through a manyToOne relationship. I was considering using a trait to include the association and its getters/setters in other entities but then I faced the issue that the mappedBy parameter couldn't be replaced correctly. If there is no way to give arguments to a trait, how else could I achieve my objective, knowing I can't extend another class. Example : /** * @ORM\OneToMany(targetEntity="

问题 I have a java class "JavaClass" with a method: boolean addAll(java.util.Collection<? extends java.lang.Integer> collection) { ... } I need to create a Scala trait MyTrait that includes this method without an implementation, something like def addAll[A<:java.lang.Integer](coll: java.util.Collection[A]) : Boolean so that an implementation class MyClass extends JavaClass with MyTrait would not be abstract. But I can't figure out any way to match the Java signature in Trait. FWIW... I'm trying to

问题 I'm trying to solve this problem : how to evaluate if a class has a specific function with CRTP. I got this as my class trait evaluation : template <typename T> class function_exists_trait { private: template <typename U> static auto test_todo(U * u) -> decltype(&U::exists) {} static auto test_todo(...) -> std::false_type {} public: static constexpr bool value{ !std::is_same<decltype(test_todo(static_cast<T*>(nullptr))), std::false_type>::value }; }; And i tested it with some trivial classes

问题 In the Scala spec, it's said that in a class template sc extends mt1, mt2, ..., mtn Each trait reference mti must denote a trait. By contrast, the superclass constructor sc normally refers to a class which is not a trait. It is possible to write a list of parents that starts with a trait reference, e.g. mt1 with …… with mtn. In that case the list of parents is implicitly extended to include the supertype of mt1 as first parent type. The new supertype must have at least one constructor that

问题 Please explain this in Scala. If I have a trait A I cannot do a val a = new A But this example trait, trait DS[-In, +Out]{def apply(i: In): Out} can have an instance of val t1 = new DS[Any, Int]{def apply(i: Any) = i.toString.toInt} How is this allowed? 回答1: Works just fine with a class body {} . val a = new A {} 回答2: What is happening is that by providing a class body you are creating an anonymous class inline that extends the trait. 来源： https://stackoverflow.com/questions/45241863/creating

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 5 years ago . I'm looking at Scala traits, but have yet to program in it, making this a hypothetical question. All of the examples that I've seen so far have concrete or abstract fields that only use val in their declaration instead of var . Can fields declared with var be used in Scala traits? TIA, Matthew 回答1: If you try it

问题 Given, from enthought.traits.api import HasTraits, Tuple, Delegate, Trait, Float,Dict,List class Foo(HasTraits): def __init__(self): super(Foo,self).__init__() self.add_trait('node',List) def _node_items_changed(self,name,old,new): print name print old print new Why do I get: >>> f = Foo() >>> f.node.append(0) node_items <undefined> <traits.trait_handlers.TraitListEvent object at 0x05BA8CF0> The documentation says I should get a list of items added/removed. What am I missing here? This is

问题 Given the following: scala> trait Foo defined trait Foo scala> case object Bip extends Foo defined module Bip scala> case object Bar extends Foo defined module Bar Is there any feature, built into Scala, that can make a Foo from a String ? example: f("Bip") === Bip f("Bar") === Bar f("...") === Exception (or maybe returns None )? 回答1: You could use macros, but it may be easier to just use simple Java reflection: namespace org.example import scala.util.control.Exception._ object Demo extends

问题 This question is maybe more about Scala than Play, but here it is: I am trying to achieve an abstraction of a repository for common DB operations. trait Entity { def id: UUID } trait Repository[T <: Entity] { val JSON_KEY_ID = "_id" def collection: JSONCollection def insert(t: T): Future[Either[String, T]] = { collection.insert(t).map(wr => if (wr.ok) Right(t) else Left(wr.getMessage())) .recover { case t => Left(t.getMessage) } } def update(t: T): Future[Either[String, T]] = { val selector =

问题 I have a trait MyTrait , and I want all trait objects &MyTrait to be comparable to each other and to nothing else. I have that now based on How to test for equality between trait objects?. The problem is that I need to use MyTraitComparable everywhere, instead of MyTrait . Is there a way to get around this? use std::any::Any; trait MyTrait {} trait MyTraitComparable: MyTrait { fn as_any(&self) -> &Any; fn equals(&self, other: &MyTraitComparable) -> bool; } impl<S: 'static + MyTrait +