How to write the type declaration of an haskell function without arguments?
2 Answers
There is no such thing as a function without arguments, that would be just a value. Sure, you can write such a declaration:
five :: Int
five = 5
It might look more like what you asked for if I make it
five' :: () -> Int
five' () = 5
but that's completely equivalent (unless you write something ridiculous like five' undefined) and superfluent1.
If what you mean is something like, in C
void scream() {
printf("Aaaah!\n");
}
then that's again not a function but an action. (C programmers do call it function, but you might better say procedure, everybody would understand.) What I said above holds pretty much the same way, you'd use
scream :: IO()
scream = putStrLn "Aaaah!"
Note that the empty () do in this case not have anything to do with not having arguments (that follows already from the absence of -> arrows), instead it means there is also no return value, it's just a "side-effect-only" action.
1Actually, it differs in one relevant way:
five is a constant applicative form, which sort of means it's memoised. If I had defined such a constant in some roundabout way (e.g. sum $ 5 : replicate 1000000 0) then the lengthy calculation would be carried out only once, even if five is evaluated multiple times during a program run. OTOH, wherever you would have written out five' (), the calculation would have been done anew.
6 Comments
danza
you can call it action or procedure, but syntactically it is just a function like all the others, and yes, in my case i actually wanted to group some side effects together, and i wanted to know how to declare the expected returned type. thanks
leftaroundabout
It is syntactically a function in almost every programming language, but not in Haskell. Here, function has a very specific meaning – basically, a lambda abstraction. If something doesn't permit you to apply it to some argument, then it's not a function. Read this blog post by Conal.
Max
So what should you use instead? How can you put a value in scope for every function?
leftaroundabout
@Max: any value that's declared on the top-level will be in scope for every function, no need to do anything. (And, of course, also for things that aren't functions...)
leftaroundabout
@Infinity: without having watched the referenced video, I assume they're talking about a point-free definition. You can define a function without explicitly mentioning any parameters, by using composition operators instead. But a function defined this way sure has parameters too, only, they're implicit in the definition. If you will,
f = (+1) is just syntactic sugar for f x = x+1: both define a function with one argument, but the former style avoids actually giving it a name. |
Since functions in Haskell are pure (their result only depends on their arguments), the equivalent of a function with no arguments is just a value. For example, one = 1.
5of typeInt, or something that really has a function type but is defined without any explicit arguments, likef = const 3?f = const 3has a type signaturef :: b -> Integer. It has thus an argument, it just ignores it