Difference between let, fun and function in F# - function

I'm learning F# and I cannot figure out what the difference between let, fun and function is, and my text book doesn't really explain that either. As an example:
let s sym = function
| V x -> Map.containsKey x sym
| A(f, es) -> Map.containsKey f sym && List.forall (s sym) es;;
Couldn't I have written this without the function keyword? Or could I have written that with fun instead of function? And why do I have to write let when I've seen some examples where you write
fun s x =
...
What's the difference really?

I guess you should really ask MSDN, but in a nutshell:
let binds a value with a symbol. The value can be a plain type like an int or a string, but it can also be a function. In FP functions are values and can be treated in the same way as those types.
fun is a keyword that introduces an anonymous function - think lambda expression if you're familiar with C#.
Those are the two important ones, in the sense that all the others usages you've seen can be thought as syntax sugar for those two. So to define a function, you can say something like this:
let myFunction =
fun firstArg secondArg ->
someOperation firstArg secondArg
And that's very clear way of saying it. You declare that you have a function and then bind it to the myFunction symbol.
But you can save yourself some typing by just conflating anonymous function declaration and binding it to a symbol with let:
let myFunction firstArg secondArg =
someOperation firstArg secondArg
What function does is a bit trickier - you combine an anonymous single-argument function declaration with a match expression, by matching on an implicit argument. So these two are equivalent:
let myFunction firstArg secondArg =
match secondArg with
| "foo" -> firstArg
| x -> x
let myFunction firstArg = function
| "foo" -> firstArg
| x -> x
If you're just starting on F#, I'd steer clear of that one. It has its uses (mainly for providing succinct higher order functions for maps/filters etc.), but results in code less readable at a glance.

These things are sort of shortcuts to each other.
The most fundamental thing is let. This keyword gives names to stuff:
let name = "stuff"
Speaking more technically, the let keyword defines an identifier and binds it to a value:
let identifier = "value"
After this, you can use words name and identifier in your program, and the compiler will know what they mean. Without the let, there wouldn't be a way to name stuff, and you'd have to always write all your stuff inline, instead of referring to chunks of it by name.
Now, values come in different flavors. There are strings "some string", there are integer numbers 42, floating point numbers 5.3, Boolean values true, and so on. One special kind of value is function. Functions are also values, in most respects similar to strings and numbers. But how do you write a function? To write a string, you use double quotes, but what about function?
Well, to write a function, you use the special word fun:
let squareFn = fun x -> x*x
Here, I used the let keyword to define an identifier squareFn, and bind that identifier to a value of the function kind. Now I can use the word squareFn in my program, and the compiler will know that whenever I use it I mean a function fun x -> x*x.
This syntax is technically sufficient, but not always convenient to write. So in order to make it shorter, the let binding takes an extra responsibility upon itself and provides a shorter way to write the above:
let squareFn x = x*x
That should do it for let vs fun.
Now, the function keyword is just a short form for fun + match. Writing function is equivalent to writing fun x -> match x with, period.
For example, the following three definitions are equivalent:
let f = fun x ->
match x with
| 0 -> "Zero"
| _ -> "Not zero"
let f x = // Using the extra convenient form of "let", as discussed above
match x with
| 0 -> "Zero"
| _ -> "Not zero"
let f = function // Using "function" instead of "fun" + "match"
| 0 -> "Zero"
| _ -> "Not zero"

Related

Unable to define a parser in Haskell: Not in scope: type variable ‘a’

I am trying to define a parser in Haskell. I am a total beginner and somehow didn't manage to find any solution to my problem at all.
For the first steps I tried to follow the instructions on the slides of a powerpoint presentation. But I constantly get the error "Not in scope: type variable ‘a’":
type Parser b = a -> [(b,a)]
item :: Parser Char
item = \inp -> case inp of
[] -> []
(x:xs) -> [(x:xs)]
error: Not in scope: type variable ‘a’
|
11 | type Parser b = a -> [(b,a)]
| ^
I don't understand the error but moreover I don't understand the first line of the code as well:
type Parser b = a -> [(b,a)]
What is this supposed to do? On the slide it just tells me that in Haskell, Parsers can be defined as functions. But that doesn't look like a function definition to me. What is "type" doing here? If it s used to specify the type, why not use "::" like in second line above? And "Parser" seems to be a data type (because we can use it in the type definition of "item"). But that doesn't make sense either.
The line derives from:
type Parser = String -> (String, Tree)
The line I used in my code snippet above is supposed to be a generalization of that.
Your help would be much appreciated. And please bear in mind that I hardly know anything about Haskell, when you write an answer :D
There is a significant difference between the type alias type T = SomeType and the type annotation t :: SomeType.
type T = Int simply states that T is just another name for the type Int. From now on, every time we use T, it will be replaced with Int by the compiler.
By contrast, t :: Int indicates that t is some value of type Int. The exact value is to be specified by an equation like t = 42.
These two concepts are very different. On one hand we have equations like T = Int and t = 42, and we can replace either side with the other side, replacing type with types and values with values. On the other hand, the annotation t :: Int states that a value has a given type, not that the value and the type are the same thing (which is nonsensical: 42 and Int have a completely different nature, a value and a type).
type Parser = String -> (String, Tree)
This correctly defines a type alias. We can make it parametric by adding a parameter:
type Parser a = String -> (String, a)
In doing so, we can not use variables in the right hand side that are not parameters, for the same reason we can not allow code like
f x = x + y -- error: y is not in scope
Hence you need to use the above Parser type, or some variation like
type Parser a = String -> [(String, a)]
By contrast, writing
type Parser a = b -> [(b, a)] -- error
would use an undeclared type b, and is an error. At best, we could have
type Parser a b = b -> [(b, a)]
which compiles. I wonder, though, is you really need to make the String type even more general than it is.
So, going back to the previous case, a possible way to make your code run is:
type Parser a = String -> [(a, String)]
item :: Parser Char
item = \inp -> case inp of
[] -> []
(x:xs) -> [(x, xs)]
Note how [(x, xs)] is indeed of type [(Char, String)], as needed.
If you really want to generalize String as well, you need to write:
type Parser a b = b -> [(b, a)]
item :: Parser Char String
item = \inp -> case inp of
[] -> []
(x:xs) -> [(xs, x)]

How does let binding work

I'm new to Haskell and I'm having troubles understanding how the let binding works in the following example:
prefixes :: [a] -> [[a]]
prefixes xs =
let prefix n = take n xs
in map prefix (range (length xs))
'take' function returns a list, so how does this get bind to 2 variables (prefix n)? Or am I totally missing the point here...
You can think of let as syntactic sugar for using an anonymous function.
let name = value in stuff is equivalent to (\name -> stuff) value. An anonymous function whose body is the expression in the in clause is applied to the expression bound to a name in the let clause.

understand syntax in the sml language

Hello I started to write in sml and I have some difficulty in understanding a particular function.
I have this function:
fun isInRow (r:int) ((x,y)) = x=r;
I would be happy to get explain to some points:
What the function accepts and what it returns.
What is the relationship between (r: int) ((x, y)).
Thanks very much !!!
The function isInRow has two arguments. The first is named r. The second is a pair (x, y). The type ascription (r: int) says that r must be an int.
This function is curried, which is a little unusual for SML. What this means roughly speaking is that it accepts arguments given separately rather than supplied as a pair.
So, the function accepts an int and a pair whose first element is an int. These are accepted as separate arguments. It returns a boolean value (the result of the comparison x = r).
A call to the function would look like this:
isInRow 3 (3, 4)
There is more to say about currying (which is kind of cool), but I hope this is enough to get you going.
In addition to what Jeffrey has said,
You don't need the extra set of parentheses:
fun isInRow (r:int) (x,y) = x=r;
You don't need to specify the type :int. If you instead write:
fun isInRow r (x,y) = x=r;
then the function's changes type from int → (int • 'a) → bool into ''a → (''a • 'b) → bool, meaning that r and x can have any type that can be compared for equality (not just int), and y can still be anything since it is still disregarded.
Polymorphic functions are one of the strengths of typed, functional languages like SML.
You could even refrain from giving y a name:
fun isInRow r (x,_) = x=r;

PolyML Functions and Types

[...] a pair of functions tofun : int -> ('a -> 'a) and fromfun : ('a -> 'a) ->
int such that (fromfun o tofun) n evaluates to n for every n : int.
Anyone able to explain to me what this is actually asking for? I'm looking for more of an explanation of that than an actual solution to this.
What this is asking for is:
1) A higher-order function tofun which when given an integer returns a polymorphic function, one which has type 'a->'a, meaning that it can be applied to values of any type, returning a value of the same type. An example of such a function is:
- fun id x = x;
val id = fn : 'a -> 'a
for example, id "cat" = "cat" and id () = (). The later value is of type unit, which is a type with only 1 value. Note that there is only 1 total function from unit to unit (namely, id or something equivalent). This underscores the difficulty with coming up with defining tofun: it returns a function of type 'a -> 'a, and other than the identity function it is hard to think of other functions. On the other hand -- such functions can fail to terminate or can raise an error and still have type 'a -> 'a.
2) fromfun is supposed to take a function of type 'a ->'a and return an integer. So e.g. fromfun id might evaluate to 0 (or if you want to get tricky it might never terminate or it might raise an error)
3) These are supposed to be inverses of each other so that, e.g. fromfun (tofun 5) needs to evaluate to 5.
Intuitively, this should be impossible in a sufficiently pure functional language. If it is possible in SML, my guess is that it would be by using some of the impure features of SML (which allow for side effects) to violate referential transparency. Or, the trick might involve raising and handling errors (which is also an impure feature of SML). If you find an answer which works in SML it would be interesting to see if it could be translated to the annoyingly pure functional language Haskell. My guess is that it wouldn't translate.
You can devise the following property:
fun prop_inverse f g n = (f o g) n = n
And with definitions for tofun and fromfun,
fun tofun n = ...
fun fromfun f = ...
You can test that they uphold the property:
val prop_test_1 =
List.all
(fn i => prop_inverse fromfun tofun i handle _ => false)
[0, ~1, 1, valOf Int.maxInt, valOf Int.minInt]
And as John suggests, those functions must be impure. I'd also go with exceptions.

How to create matching pattern on a pair of functions in haskell [duplicate]

Imagine I have a custom type and two functions:
type MyType = Int -> Bool
f1 :: MyType -> Int
f3 :: MyType -> MyType -> MyType
I tried to pattern match as follows:
f1 (f3 a b i) = 1
But it failed with error: Parse error in pattern: f1. What is the proper way to do the above?? Basically, I want to know how many f3 is there (as a and b maybe f3 or some other functions).
You can't pattern match on a function. For (almost) any given function, there are an infinite number of ways to define the same function. And it turns out to be mathematically impossible for a computer to always be able to say whether a given definition expresses the same function as another definition. This also means that Haskell would be unable to reliably tell whether a function matches a pattern; so the language simply doesn't allow it.
A pattern must be either a single variable or a constructor applied to some other patterns. Remembering that constructor start with upper case letters and variables start with lower case letters, your pattern f3 a n i is invalid; the "head" of the pattern f3 is a variable, but it's also applied to a, n, and i. That's the error message you're getting.
Since functions don't have constructors, it follows that the only pattern that can match a function is a single variable; that matches all functions (of the right type to be passed to the pattern, anyway). That's how Haskell enforces the "no pattern matching against functions" rule. Basically, in a higher order function there's no way to tell anything at all about the function you've been given except to apply it to something and see what it does.
The function f1 has type MyType -> Int. This is equivalent to (Int -> Bool) -> Int. So it takes a single function argument of type Int -> Bool. I would expect an equation for f1 to look like:
f1 f = ...
You don't need to "check" whether it's an Int -> Bool function by pattern matching; the type guarantees that it will be.
You can't tell which one it is; but that's generally the whole point of taking a function as an argument (so that the caller can pick any function they like knowing that you'll use them all the same way).
I'm not sure what you mean by "I want to know how many f3 is there". f1 always receives a single function, and f3 is not a function of the right type to be passed to f1 at all (it's a MyType -> MyType -> MyType, not a MyType).
Once a function has been applied its syntactic form is lost. There is now way, should I provide you 2 + 3 to distinguish what you get from just 5. It could have arisen from 2 + 3, or 3 + 2, or the mere constant 5.
If you need to capture syntactic structure then you need to work with syntactic structure.
data Exp = I Int | Plus Exp Exp
justFive :: Exp
justFive = I 5
twoPlusThree :: Exp
twoPlusThree = I 2 `Plus` I 3
threePlusTwo :: Exp
threePlusTwo = I 2 `Plus` I 3
Here the data type Exp captures numeric expressions and we can pattern match upon them:
isTwoPlusThree :: Exp -> Bool
isTwoPlusThree (Plus (I 2) (I 3)) = True
isTwoPlusThree _ = False
But wait, why am I distinguishing between "constructors" which I can pattern match on and.... "other syntax" which I cannot?
Essentially, constructors are inert. The behavior of Plus x y is... to do nothing at all, to merely remain as a box with two slots called "Plus _ _" and plug the two slots with the values represented by x and y.
On the other hand, function application is the furthest thing from inert! When you apply an expression to a function that function (\x -> ...) replaces the xes within its body with the applied value. This dynamic reduction behavior means that there is no way to get a hold of "function applications". They vanish into thing air as soon as you look at them.