I'm new to scala and i'm trying to understand the real differences between those 3 syntaxes of code :
//first code
def add(x:Int, y:Int) = {x+y}
//second code
val add2 = (x:Int,y:Int) => x+y
//third code
def add3 = (x:Int,y:Int) => x+y
I can approximatively see the differences but I don't know which one should I use depends on the context.
Is anyone have concrete examples ?
Thanks a lot !
This is a FAQ.
But, given you also asked when to use each I guess it is worth adding a little bit over that.
First, let's explain what does each mean:
Is a method of two arguments, both are Int and it returns another Int; as such, it is not a value.
Is a function of two arguments, both are Ints and it returns another Int; as such, it is a value whose type is Function2[Int, Int, Int] (commonly known as (Int, Int) => Int).
Is a method of zero arguments, and it returns a function (Int, Int) => Int
In general always use the first one, since it is more powerful (Scala 3 will reduce the differences between methods and functions, but still), for most people the syntax is more clear, it should be more efficient. And, even if you are going to use it as a function, for example for a map, then eta-expansion will take care of that.
Use the second one when you are absolutely sure you need it as a function, for example you know you will be using things like andThen with other functions; this is not very common.
Never use the third one since it would be creating a new object every time you call it, to then discard it after its use (which would be very inefficient) and it should be the same as using val instead of def; the only "valid" reason for a method of no arguments that returns a function, would be that the function is always different but that would imply a side-effect that is discouraged.
What constitutes a value in pure functional programming?
I am asking myself these questions after seeing a sentence:
Task(or IO) has a constructor that captures side-effects as values.
Is a function a value?
If so, what does it mean when equating two functions: assert(f == g). For two functions that are equivalent but defined separately => f != g, why don't they work as 1 == 1?
Is an object with methods a value? (for example IO { println("") })
Is an object with setter methods and mutable state a value?
Is an object with mutable state which works as a state machine a value?
How do we test whether something is a value? Is immutability a sufficient condition?
UPDATE:
I'm using Scala.
I'll try to explain what a value is by contrasting it with things that are not values.
Roughly speaking, values are structures produced by the process of evaluation which correspond to terms that cannot be simplified any further.
Terms
First, what are terms? Terms are syntactic structures that can be evaluated. Admittedly, this is a bit circular, so let's look at a few examples:
Constant literals are terms:
42
Functions applied to other terms are terms:
atan2(123, 456 + 789)
Function literals are terms
(x: Int) => x * x
Constructor invocations are terms:
Option(42)
Contrast this to:
Class declarations / definitions are not terms:
case class Foo(bar: Int)
that is, you cannot write
val x = (case class Foo(bar: Int))
this would be illegal.
Likewise, trait and type definitions are not terms:
type Bar = Int
sealed trait Baz
Unlike function literals, method definitions are not terms:
def foo(x: Int) = x * x
for example:
val x = (a: Int) => a * 2 // function literal, ok
val y = (def foo(a: Int): Int = a * 2) // no, not a term
Package declarations and import statements are not terms:
import foo.bar.baz._ // ok
List(package foo, import bar) // no
Normal forms, values
Now, when it is hopefully somewhat clearer what a term is, what was meant by "cannot be simplified any further*? In idealized functional programming languages, you can define what a normal form, or rather weak head normal form is. Essentially, a term is in a (wh-) normal form if no reduction rules can be applied to the term to make it any simpler. Again, a few examples:
This is a term, but it's not in normal form, because it can be reduced to 42:
40 + 2
This is not in weak head normal form:
((x: Int) => x * 2)(3)
because we can further evaluate it to 6.
This lambda is in weak head normal form (it's stuck, because the computation cannot proceed until an x is supplied):
(x: Int) => x * 42
This is not in normal form, because it can be simplified further:
42 :: List(10 + 20, 20 + 30)
This is in normal form, no further simplifications possible:
List(42, 30, 50)
Thus,
42,
(x: Int) => x * 42,
List(42, 30, 50)
are values, whereas
40 + 2,
((x: Int) => x * 2)(3),
42 :: List(10 + 20, 20 + 30)
are not values, but merely non-normalized terms that can be further simplified.
Examples and non-examples
I'll just go through your list of sub-questions one-by-one:
Is a function a value
Yes, things like (x: T1, ..., xn: Tn) => body are considered to be stuck terms in WHNF, in functional languages they can actually be represented, so they are values.
If so, what does it mean when equating two functions: assert(f == g) for two functions that are equivalent but defined separately => f != g, why don't they work as 1 == 1?
Function extensionality is somewhat unrelated to the question whether something is a value or not. In the above "definition by example", I talked only about the shape of the terms, not about the existence / non-existence of some computable relations defined on those terms. The sad fact is that you can't even really determine whether a lambda-expression actually represents a function (i.e. whether it terminates for all inputs), and it is also known that there cannot be an algorithm that could determine whether two functions produce the same output for all inputs (i.e. are extensionally equal).
Is an object with methods a value? (for example IO { println("") })
Not quite clear what you're asking here. Objects don't have methods. Classes have methods. If you mean method invocations, then, no, they are terms that can be further simplified (by actually running the method), so they are not values.
Is an object with setter methods and mutable state a value?
Is an object with mutable state which works as a state machine a value?
There is no such thing in pure functional programming.
What constitutes a value in pure functional programming?
Background
In pure functional programming there is no mutation. Hence, code such as
case class C(x: Int)
val a = C(42)
val b = C(42)
would become equivalent to
case class C(x: Int)
val a = C(42)
val b = a
since, in pure functional programming, if a.x == b.x, then we would have a == b. That is, a == b would be implemented comparing the values inside.
However, Scala is not pure, since it allows mutation, like Java. In such case, we do NOT have the equivalence between the two snippets above, when we declare case class C(var x: Int). Indeed, performing a.x += 1 afterwords does not affect b.x in the first snippet, but does in the second one, where a and b point to the same object. In such case, it is useful to have a comparison a == b which compares the object references, rather than its inner integer value.
When using case class C(x: Int), Scala comparisons a == b behave closer to pure functional programming, comparing the integers values. With regular (non case) classes, Scala instead compares object references breaking the equivalence between the two snippets. But, again, Scala is not pure. By comparison, in Haskell
data C = C Int deriving (Eq)
a = C 42
b = C 42
is indeed equivalent to
data C = C Int deriving (Eq)
a = C 42
b = a
since there are no "references" or "object identities" in Haskell. Note that the Haskell implementation likely will allocate two "objects" in the first snippet, and only one object in the second one, but since there is no way to tell them apart inside Haskell, the program output will be the same.
Answer
Is a function a value ? (then what it means when equating two function: assert(f==g). For two function that is equivalent but defined separately => f!=g, why not they work like 1==1)
Yes, functions are values in pure functional programming.
Above, when you mention "function that is equivalent but defined separately", you are assuming that we can compare the "references" or "object identities" for these two functions. In pure functional programming we can not.
Pure functional programming should compare functions making f == g equivalent to f x == g x for all possible arguments x. This is feasible when there is only a few values for x, e.g. if f,g :: Bool -> Int we only need to check x=True, x=False. For functions having infinite domains, this is much harder. For instance, if f,g :: String -> Int we can not check infinitely many strings.
Theoretical computer science (computability theory) also proved that there is no algorithm to compare two functions String -> Int, not even an inefficient algorithm, not even if we have access to the source code of the two functions. For this mathematical reason, we must accept that functions are values that can not be compared. In Haskell, we express this through the Eq typeclass, stating that almost all the standard types are comparable, functions being the exception.
Is an object with methods a value ? (for example, IO{println("")})
Yes. Roughly speaking, "everything is a value", including IO actions.
Is an object with setter methods and mutable states a value ?
Is an object with mutable states and works as a state machine a value ?
There is no mutable state in pure functional programming.
At best, the setters can produce a "new" object with the modified fields.
And yes, the object would be a value.
How do we test if it is a value, is that immutable can be a sufficient condition to be a value ?
In pure functional programming, we can only have immutable data.
In impure functional programming, I think we can call most immutable objects "values", when we do not compare object references. If the "immutable" object contains a reference to a mutable object, e.g.
case class D(var x: Int)
case class C(c: C)
val a = C(D(42))
then things are more tricky. I guess we could still call a "immutable", since we can not alter a.c, but we should be careful since a.c.x can be mutated.
Depending on the intent, I think that some would not call a immutable. I would not consider a to be a value.
To make things more muddy, in impure programming, there are objects which use mutation to present a "pure" interface in an efficient way. For instance one can write a pure function that, before returning, stores its result in a cache. When called again on the same argument, it will return the previously computed result
(this is usually called memoization). Here, mutation happens, but it is not observable from outside, where at most we can observe a faster implementation. In this case, we can simply pretend the that function is pure (even if it performs mutation) and consider it a "value".
The contrast with imperative languages is stark. In inperitive languages, like Python, the output of a function is directed. It can be assigned to a variable, explicitly returned, printed or written to a file.
When I compose a function in Haskell, I never consider output. I never use "return" Everything has "a" value. This is called "symbolic" programming. By "everything", is meant "symbols". Like human language, the nouns and verbs represent something. That something is their value. The "value" of "Pete" is Pete. The name "Pete" is not Pete but is a representation of Pete, the person. The same is true of functional programming. The best analogy is math or logic When you do pages of calculations, do you direct the output of each function? You even "assign" variables to be replaced by their "value" in functions or expressions.
Values are
Immutable/Timeless
Anonymous
Semantically Transparent
What is the value of 42? 42. What is the "value" of new Date()? Date object at 0x3fa89c3. What is the identity of 42? 42. What is the identity of new Date()? As we saw in the previous example, it's the thing that lives at the place. It may have many different "values" in different contexts but it has only one identity. OTOH, 42 is sufficient unto itself. It's semantically meaningless to ask where 42 lives in the system. What is the semantic meaning of 42? Magnitude of 42. What is the semantic meaning of new Foo()? Who knows.
I would add a fourth criterion (see this in some contexts in the wild but not others) which is: values are language agnostic (I'm not certain the first 3 are sufficient to guarantee this nor that such a rule is entirely consistent with most people's intuition of what value means).
Values are things that
functions can take as inputs and return as outputs, that is, can be computed, and
are members of a type, that is, elements of some set, and
can be bound to a variable, that is, can be named.
First point is really the crucial test whether something is a value. Perhaps the word value, due to conditioning, might immediately make us think of just numbers, but the concept is very general. Essentially anything we can give to and get out of a function can be considered a value. Numbers, strings, booleans, instances of classes, functions themselves, predicates, and even types themselves, can be inputs and outputs of functions, and thus are values.
IO monad is a great example of how general this concept is. When we say IO monad models side-effects as values, we mean a function can take a side-effect (say println) as input and return as output. IO(println(...)) separates the idea of the effect of an action of println from the actual execution of an action, and allows these effects to be considered as first class values that can be computed with using the same language facilities as for any other values such as numbers.
More explicitly, can this code produce any errors in any scenrios:
def foreach[U](f :Int=>U) = f.asInstanceOf[Int=>Unit](1)
I know it works, and I have a vague idea why: any function, as an instance of a generic type, must define an erased version of apply and jvm performs type check only when the object is actually to be returned to a code where it had a concrete type (often miles away). So, in theory, as long as I never look at the returned value, I should be safe. I don't have an enough low-level understandings of java byte code, let alone scalac, to have any certainty about it.
Why would I want to do it? Look at the following example:
val b = new mutable.Buffer[Int]
val ints = Seq(1, 2, 3, 4)
ints foreach { b += _ }
It's a typical scala construct, as far as imperative style can be typical. foreach in this example takes an Int as an argument, and as scalac knows it to be an Int, it will create a closure with a specialized apply(x :Int). Unfortunately, its return type in this case is a mutable.Buffer[Int], which is an AnyRef. As far as I was able to see, scalac will never invoke a specialized apply providing an AnyVal argument if the result is an AnyRef (and vice versa). This means, that even if the caller applies the function to Int, underneath the function will box the argument and invoke the erased variant. Here of course it doesn't matter as they are boxed within the List anyway, but I'm talking about the principle.
For this reason I prefer to define this type of method as foreach(f :X=>Unit), rather than foreach[O](f: X=>O) as it is in TraversableOnce. If the input sequence in the example had such a signature, everything would compile just as fine, and the compiler would ignore the actual type of the expression and generate a function with Unit return type, which - when applied to an unboxed Int - would invoke directly void apply(Int x), without boxing.
The problem arises with interoperability - sometimes I need to call a method expecting a function with a Unit return type and all I have is a generic function returning Odin knows what. Of course, I could just write f(_) to box it in another function object instead of passing it directly, but it to large extent makes the whole optimisation of small tight loops moot.
I'm a beginner of Scala who is struggling with Scala syntax.
I got the line of code from https://www.tutorialspoint.com/scala/higher_order_functions.htm.
I know (x: A) is an argument of layout function
( which means argument x of Type A)
But what is [A] between layout and (x: A)?
I've been googling scala function syntax, couldn't find it.
def layout[A](x: A) = "[" + x.toString() + "]"
It's a type parameter, meaning that the method is parameterised (some also say "generic"). Without it, compiler would think that x: A denotes a variable of some concrete type A, and when it wouldn't find any such type it would report a compile error.
This is a fairly common thing in statically typed languages; for example, Java has the same thing, only syntax is <A>.
Parameterized methods have rules where the types can occur which involve concepts of covariance and contravariance, denoted as [+A] and [-A]. Variance is definitely not in the scope of this question and is probably too much for you too handle right now, but it's an important concept so I figured I'd just mention it, at least to let you know what those plus and minus signs mean when you see them (and you will).
Also, type parameters can be upper or lower bounded, denoted as [A <: SomeType] and [A >: SomeType]. This means that generic parameter needs to be a subtype/supertype of another type, in this case a made-up type SomeType.
There are even more constructs that contribute extra information about the type (e.g. context bounds, denoted as [A : Foo], used for typeclass mechanism), but you'll learn about those later.
This means that the method is using a generic type as its parameter. Every type you pass that has the definition for .toString could be passed through layout.
For example, you could pass both int and string arguments to layout, since you could call .toString on both of them.
val i = 1
val s = "hi"
layout(i) // would give "[1]"
layout(s) // would give "[hi]"
Without the gereric parameter, for this example you would have to write two definitions for layout: one that accepts integers as param, and one that accepts string as param. Even worse: every time you need another type you'd have to write another definition that accepts it.
Take a look at this example here and you'll understand it better.
I also recomend you to take a look at generic classes here.
A is a type parameter. Rather than being a data type itself (Ex. case class A), it is generic to allow any data type to be accepted by the function. So both of these will work:
layout(123f) [Float datatype] will output: "[123]"
layout("hello world") [String datatype] will output: "[hello world]"
Hence, whichever datatype is passed, the function will allow. These type parameters can also specify rules. These are called contravariance and covariance. Read more about them here!
In Scala, if I define a method called apply in a class or a top-level object, that method will be called whenever I append a pair a parentheses to an instance of that class, and put the appropriate arguments for apply() in between them. For example:
class Foo(x: Int) {
def apply(y: Int) = {
x*x + y*y
}
}
val f = new Foo(3)
f(4) // returns 25
So basically, object(args) is just syntactic sugar for object.apply(args).
How does Scala do this conversion?
Is there a globally defined implicit conversion going on here, similar to the implicit type conversions in the Predef object (but different in kind)? Or is it some deeper magic? I ask because it seems like Scala strongly favors consistent application of a smaller set of rules, rather than many rules with many exceptions. This initially seems like an exception to me.
I don't think there's anything deeper going on than what you have originally said: it's just syntactic sugar whereby the compiler converts f(a) into f.apply(a) as a special syntax case.
This might seem like a specific rule, but only a few of these (for example, with update) allows for DSL-like constructs and libraries.
It is actually the other way around, an object or class with an apply method is the normal case and a function is way to construct implicitly an object of the same name with an apply method. Actually every function you define is an subobject of the Functionn trait (n is the number of arguments).
Refer to section 6.6:Function Applications of the Scala Language Specification for more information of the topic.
I ask because it seems like Scala strongly favors consistent application of a smaller set of rules, rather than many rules with many exceptions.
Yes. And this rule belongs to this smaller set.