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What is the apply function in Scala?

I never understood it from the contrived unmarshalling and verbing nouns ( an AddTwo class has an apply that adds two!) examples.

I understand that it's syntactic sugar, so (I deduced from context) it must have been designed to make some code more intuitive.

What meaning does a class with an apply function give? What is it used for, and what purposes does it make code better (unmarshalling, verbing nouns etc)?

how does it help when used in a companion object?

Even a quick googling brings a lots of nice articles. Here is one: jackcoughonsoftware.blogspot.com/2009/01/…
actually it is the same as a constructor in Java/C++ but can return the value and has 'apply' name
It's a method, not a function. The distinction between methods and functions is very important in Scala.
To elaborate on Zoidberg, "Methods are just functions that can access the state of the class" twitter.github.io/scala_school/basics.html In general this applies to more than Scala stackoverflow.com/questions/155609/…

M
Muhammad Hewedy

Mathematicians have their own little funny ways, so instead of saying "then we call function f passing it x as a parameter" as we programmers would say, they talk about "applying function f to its argument x".

In mathematics and computer science, Apply is a function that applies functions to arguments. Wikipedia

apply serves the purpose of closing the gap between Object-Oriented and Functional paradigms in Scala. Every function in Scala can be represented as an object. Every function also has an OO type: for instance, a function that takes an Int parameter and returns an Int will have OO type of Function1[Int,Int].

 // define a function in scala
 (x:Int) => x + 1

 // assign an object representing the function to a variable
 val f = (x:Int) => x + 1

Since everything is an object in Scala f can now be treated as a reference to Function1[Int,Int] object. For example, we can call toString method inherited from Any, that would have been impossible for a pure function, because functions don't have methods:

  f.toString

Or we could define another Function1[Int,Int] object by calling compose method on f and chaining two different functions together:

 val f2 = f.compose((x:Int) => x - 1)

Now if we want to actually execute the function, or as mathematician say "apply a function to its arguments" we would call the apply method on the Function1[Int,Int] object:

 f2.apply(2)

Writing f.apply(args) every time you want to execute a function represented as an object is the Object-Oriented way, but would add a lot of clutter to the code without adding much additional information and it would be nice to be able to use more standard notation, such as f(args). That's where Scala compiler steps in and whenever we have a reference f to a function object and write f (args) to apply arguments to the represented function the compiler silently expands f (args) to the object method call f.apply (args).

Every function in Scala can be treated as an object and it works the other way too - every object can be treated as a function, provided it has the apply method. Such objects can be used in the function notation:

// we will be able to use this object as a function, as well as an object
object Foo {
  var y = 5
  def apply (x: Int) = x + y
}


Foo (1) // using Foo object in function notation 

There are many usage cases when we would want to treat an object as a function. The most common scenario is a factory pattern. Instead of adding clutter to the code using a factory method we can apply object to a set of arguments to create a new instance of an associated class:

List(1,2,3) // same as List.apply(1,2,3) but less clutter, functional notation

// the way the factory method invocation would have looked
// in other languages with OO notation - needless clutter
List.instanceOf(1,2,3) 

So apply method is just a handy way of closing the gap between functions and objects in Scala.


how would you add custom variable type to an object. AFAIK it's not possible. Here is an example: You have this class class Average[YType](yZeroValue:YType). How do you pass YType from it's object since objects can't take type params?
Types in Scala can usually be inferred based on the arguments, but if not you can supply them via the square brackets. so when instantiating an Average object you could say "val avg = new Average[Int](0)"
Case classes are worth a mention here. Case classes conveniently, automagically and invisibly adds apply and unapply methods to a companion object of the class (among other things). So defining a class with just one line like this case class Car(make:String, model: String, year: Short, color: String) makes it possible to produce new objects of the Car class by just: val myCar = Car("VW","Golf",1999,"Blue"). This is shorthand for Car.apply(...)
every object can be treated as a function, provided it has the apply method - so much now makes sense.
T
Tristan Reid

It comes from the idea that you often want to apply something to an object. The more accurate example is the one of factories. When you have a factory, you want to apply parameter to it to create an object.

Scala guys thought that, as it occurs in many situation, it could be nice to have a shortcut to call apply. Thus when you give parameters directly to an object, it's desugared as if you pass these parameters to the apply function of that object:

class MyAdder(x: Int) {
  def apply(y: Int) = x + y
}

val adder = new MyAdder(2)
val result = adder(4) // equivalent to x.apply(4)

It's often use in companion object, to provide a nice factory method for a class or a trait, here is an example:

trait A {
  val x: Int
  def myComplexStrategy: Int
}

object A {
  def apply(x: Int): A = new MyA(x)

  private class MyA(val x: Int) extends A {
    val myComplexStrategy = 42
  }
}

From the scala standard library, you might look at how scala.collection.Seq is implemented: Seq is a trait, thus new Seq(1, 2) won't compile but thanks to companion object and apply, you can call Seq(1, 2) and the implementation is chosen by the companion object.


Could the equivalent of apply be accomplished using Implicits as well?
s
sdinesh94

Here is a small example for those who want to peruse quickly

 object ApplyExample01 extends App {


  class Greeter1(var message: String) {
    println("A greeter-1 is being instantiated with message " + message)


  }

  class Greeter2 {


    def apply(message: String) = {
      println("A greeter-2 is being instantiated with message " + message)
    }
  }

  val g1: Greeter1 = new Greeter1("hello")
  val g2: Greeter2 = new Greeter2()

  g2("world")


} 

output A greeter-1 is being instantiated with message hello A greeter-2 is being instantiated with message world


Is the message for g2 correct? Is that actually happening at instantiation, or is it in fact after instantiation (even if it's lazily instantiated)?
T
Tomáš Růžička

TLDR for people comming from c++

It's just overloaded operator of ( ) parentheses

So in scala:

class X {
   def apply(param1: Int, param2: Int, param3: Int) : Int = {
     // Do something
   }
}

Is same as this in c++:

class X {
   int operator()(int param1, int param2, int param3) {
      // do something
   }
};

j
jiawei hu

1 - Treat functions as objects.

2 - The apply method is similar to __call __ in Python, which allows you to use an instance of a given class as a function.


Ö
Özgür

To put it crudely,

You can just see it as custom ()operator. If a class X has an apply() method, whenever you call X() you will be calling the apply() method.


s
seanhalle

The apply method is what turns an object into a function. The desire is to be able to use function syntax, such as:

f(args)

But Scala has both functional and object oriented syntax. One or the other needs to be the base of the language. Scala (for a variety of reasons) chooses object oriented as the base form of the language. That means that any function syntax has to be translated into object oriented syntax.

That is where apply comes in. Any object that has the apply method can be used with the syntax:

f(args)

The scala infrastructure then translates that into

f.apply(args)

f.apply(args) has correct object oriented syntax. Doing this translation would not be possible if the object had no apply method!

In short, having the apply method in an object is what allows Scala to turn the syntax: object(args) into the syntax: object.apply(args). And object.apply(args) is in the form that can then execute.

FYI, this implies that all functions in scala are objects. And it also implies that having the apply method is what makes an object a function!

See the accepted answer for more insight into just how a function is an object, and the tricks that can be played as a result.