In this article, this function
let adderGenerator numberToAdd = (+) numberToAdd
has this type signature
int -> (int -> int)
However, when I create this function, the inferred type signature is
int -> int -> int
Are these type signatures different?
If not, how can I annotate my adderGenerator so that it has the first type signature?
Edit 1
This is the best I could come up with so far:
type Generator = int -> (int -> int)
let adderGenerator: Generator = (+)
I don't like how it makes adderGenerator adopt a point-free style.
Edit 2
This seems to work:
let adderGenerator numberToAdd : (int -> int) = (+) numberToAdd
The type signatures are the same, the parentheses just help indicate that the body of the function results in a function being returned.
These three forms with the full type annotated are equivalent:
let adderGenerator : int -> int -> int =
(+)
let adderGenerator (numberToAdd : int) : int -> int =
(+) numberToAdd
let adderGenerator : int -> int -> int =
fun numberToAdd -> (+) numberToAdd
The first form is just an alias for + and simply reiterates its type.
The second form is a function declaration that annotates the type of the argument and return type separately.
The third form is a variable assigned an anonymous function.
Only making one argument explicit seems a bit odd to me, however. You can avoid partial application entirely by making both arguments explicit:
let adderGenerator (a : int) (b : int) : int =
(+) a b
let adderGenerator : int -> int -> int =
fun a b -> (+) a b
Related
The function below compiles and runs fine.
I do not intend to create them like this (without parameter names).
However because it lets me create such a function, I am curious how I can access them within the function.
let testfunction string int =
printfn "Inside Method"
testfunction "a" 5
Here is the F# interactive output
Inside Method
val testfunction: string: 'a -> int: 'b -> unit
val it: unit = ()
Thank you for your help in advance.
Actually, string and int are the parameter names, so you access them like any other parameters:
let testfunction string int =
printfn $"string is {string}"
printfn $"int is {int}"
testfunction "a" 5
Output is:
string is a
int is 5
To reduce confusion, you could declare the function like this:
let testfunction (string : string) (int : int) =
printfn $"string is {string}"
printfn $"int is {int}"
I am using FSI, how do I go from bool to float?
have: let sigF 1 = 1=1 gets: val sigF : int -> bool
need the function that has the signature: val sigF : int -> bool -> float -> string
can you not convert bool to float?
are there any resources on understanding signatures?, I cant find any examples where bool goes to anything else, do i need to understand "currying" better? (I am completely new to functional programming)
Unless this is a puzzle or a challenge, you can assume that i:int -> b:bool -> f:float -> string is the same as int -> bool -> float -> string. The only difference is that the former also includes names of the parameters of the function - which is just extra information that you can ignore and that does not change the meaning much.
You get the parameter names if you do something like this:
> let sigF (i:int) (b:bool) (f:float) = "";;
val sigF : i:int -> b:bool -> f:float -> string
The ingineous solution from Michael avoids naming the parameters by instead using patterns (matching against concrete values), which gives you the right signature, but also a plenty of warning because the function will fail if called with any other value as argument:
> let sigF 7 true 0.3 = "done";;
warning FS0025: Incomplete pattern matches on this expression.
For example, the value '0.0' may indicate a case not covered by the pattern(s).
warning FS0025: Incomplete pattern matches on this expression.
For example, the value 'false' may indicate a case not covered by the pattern(s).
warning FS0025: Incomplete pattern matches on this expression.
For example, the value '0' may indicate a case not covered by the pattern(s).
val sigF : int -> bool -> float -> string
Another solution that gives you the right signature, but without the warnings is to use _ pattern with a type annotation - this says that you are ignoring the argumnet, but it gives an explicit type to it:
> let sigF (_:int) (_:bool) (_:float) = "";;
val sigF : int -> bool -> float -> string
let f 7 true 0.3 = "done";;
7 true and .03 (all imply function inputs) while after the = is an output
val f : int -> bool -> float -> string
A function having the signature val sigF : int -> bool -> float -> string could be for example let sigF (i:int) (b:bool) (f:float) = "". The last in the row of -> symbols is the output, all others are inputs.
Got it from Here
The Ultimate Closure
Finally, for the ultra parsimonious there is the following, without a byte wasted.
let testEquality9 : (Int, Int) -> Bool = (==)
Functions decalred with the func keyword are just closures with names. == is an example of one such named function. It takes 2 Int arguements, and returns a Bool telling you if they're equal. Its type is (Int, Int) -> Bool
testEquality9 is a closure, with the type (Int, Int) -> Bool. To it, the closure of the == function is assigned.
It can be called like this:
testEquality9(1, 2) // false
testEquality9(1, 1) // true
The key thing to draw from this is that functions are really just closures, so they can be used everywhere closures can be used.
For example, if you wanted to sort an array of Ints, you could use:
let ints = [3, 1, 4, 2]
let sorted = ints.sort{$0 < $1}
The sort(_:) method takes a closure that's of type (Int, Int) -> Bool. Our closure {$0 < $1} takes 2 Int params, and returns a Bool. So it fits that signiture.
However, we can make this code shorter. Because the < operator's function already has type (Int, Int) -> Bool, we can write this:
let sorted = ints.sort(<)
This passes the function (named closure) < in directly, without explicitly making our own closure to wrap around it.
This is not actually a closure, it's the equality operator that compares two integers stored into a variable.
Every operator is defined using a function and that function can be assigned to a variable. There is nothing else to it.
Operator Overloading:
func == (i : Int, j: Int) -> Bool {
return i == j
}
Should be equivalent to that.
As the others said, it's the abbreviation form of:
let testEquality9: (Int, Int) -> Bool = { (a: Int, b: Int) -> Bool in return a == b }
Reading from right to left, it creates a function that compares two Ints and assigns it to the constant testEquality9.
You need to mentally separate the 3 pieces:
The constant name:
let testEquality9
The constant type (it's a function type):
(Int, Int) -> Bool
And the value assigned to the constant:
(==)
OR, the long version:
{ (a: Int, b: Int) -> Bool in return a == b }
Enjoy Swift :)
Is there a way to name the function arguments in a constructor?
type UnnamedInCtor(foo: string -> string -> bool) =
member this.Foo: string -> string -> bool = foo
member this.Bar: a:string -> b:string -> bool = foo
member this.Fizz = foo
//Does not compile
type NamedInCtor(foo: a:string -> b:string -> bool) =
member this.Foo: string -> string -> bool = foo
member this.Bar: a:string -> b:string -> bool = foo
member this.Fizz = foo
I think that it's impossible in F#, however you can use type abbreviations if you want to document what foo represents:
// Compiles
type aToBToC = string -> string -> bool
type NamedInCtor(foo: aToBToC) =
member this.Foo: string -> string -> bool = foo
member this.Bar: a:string -> b:string -> bool = foo
member this.Fizz = foo
You would need to de-curry the function in your constructor:
type NamedInCtor(a, b) =
member this.Foo: string -> string -> bool = a b
member this.Bar: string -> string -> bool = a b
member this.Fizz = a b
Note that a and b are implicitly typed here. You should trust the compiler to do this as much as possible, because it makes your code much more readable.
Remember, functions are first class types and traditional objects are discouraged. What you're asking is essentially "can I name and access some arbitrary subset of this type?" The answer to that is no. If you want that behavior, then you must structure your functions to request it.
I have a question regarding two types int -> (int -> int) and (int -> int) -> int. What is the difference between these two?
I read here somewhere that the first one could be interpreted as a function that that takes an int and returns a function that takes and int and returns an int, which is the same as int ->int -> int. Is this right?
The difference between int -> (int -> int) and int -> int -> int is negligible (you would really only care if you had to interop with another language). These are curried functions which effectively take two integers and return an integer (for example, the built-in addition operator has a type along these lines).
However, both of these are very different from (int -> int) -> int, which takes a function from integers to integers and returns an integer. As others have mentioned, the only simple example of such a function would apply the given function to a specific value (e.g. fun (f:int->int) -> f 13).
Good question, I suspect that functionally there's not much difference. I spent a while fiddling around trying to work out how to even get those two signatures. (John Palmer's let fun1 i = fun j -> i + j just gives int -> int -> int for me)
let add a b = a + b // int -> int -> int
let inc a = add a // int -> (int -> int)
I couldn't think of a non-contrived example for the second signature:
let foo fn : int -> fn 1 // (int -> int) -> int
(This doesn't directly answer your question, but may give someone else something to chew on)
I dont really know F#, but it seems quite intuitive (many other language have the same notions).
Is a function taking an int, and returning a function that takes an int and returns an int
Is a function taking a function that takes an int and returns an int, and returning an int
The first would be kind of like curried function.
So lets see what implementations would look like:
first
let func1 i = fun j -> i+j
this has a signature of int -> (int -> int)
note that func1 2 3 shouldn't work
EDIT: It turns out that due to the way associativity works this is actually fine
but this is different to
let func2 i j= i+j
which has a type of int -> int -> int
here func2 1 2 is fine
the other example we can create like this:
let func4 a = a 1
this will have a signature of ('t->int) -> u picking a concrete a will work. Note that func4 1 2 will definitely not compile