Re: [Haskell-beginners] map type explanation

On Fri, 18 Dec 2020, Lawrence Bottorff wrote:

So in effect

a -> b -> [a] -> [b]

wants to be, would be

a -> (b -> ([a] -> [b]))

One difficulty in learning New Crazy Type Theory is that, by tradition, parentheses are suppressed so thoroughly that a beginner sometimes finds it hard to correctly parse the surface notation. Let us suppose we have a thing called "a". Another thing is called "b". a and b are, for the moment, just things. a and b are things of the same sort: they are both 'types'. In type theory, like in group theory, or ring theory, we have various algebras, with operations. In ring theory we ring elements, and operations: +, -, 0, 1, *. A particular ring, call the ring, say "Example" is given by the following data: 1. An underlying set, call this set Substrate_of_Example. 2. A specified element of Substrate_of_Example, the 0. 3. A specified element of Substrate_of_Example, the 1. 4. A specified everywhere defined single valued function, the +: S_E x S_E -> S_E 5. A specified everywhere defined single valued function, the *: S_E x S_E -> S_E 6. 4. A specified everywhere defined single valued function, the -: S_E -> S_E + and * take two inputs and have one output. - takes one input and has one output. By the definition of the concept "ring", the ring substrate S_E and the two constants, 0 and 1, and the three operations, +, *, and -, must satisfy certain laws. If case these laws are satisfied, we say that the data give us a uniquely defined instance of the concept ring. The sub-field of mathematics Ring Theory studies such things, that is, rings. Similarly Type Theory studies "type systems". The definition of a type system is not as tightly defined as the concept of a ring, but again, at least in the case of the Haskell_Type_System, we have these things: 1. An underlying set, call this set Substrate_of_Haskell_Type_System. This set is the set of types. For convenience we write "T" for "Substrate_of_Haskell_Type_System". 2. An everywhere defined single valued function, the hom operator =>: T x T -> T (We have written "=>" rather than "->" to the left of the ":" above to avoid confusion with the "->" in the above line.) 3. An everywhere defined single valued function, the list operator []: T -> T. (Note [] is not the list operator of, say, Common Lisp, or Scheme, or Haskell. [] operates on types, not on, ah, things which are not types, but rather have types. Oi.) A particular type system such as HTS is given by the above data, and the system is a type system, because just as for rings, certain laws are required to be obeyed. (Usually a type system has more operations than the above two.) Let us suppose we have before us the ring of integers Z. Here is an expression in the language of rings, which language applies to Z, because Z is a ring: (((1 + 1) * (0 + 1)) + ((- 1) * (1 + 1))) The notation is un-ambiguous. We know what it means, and we can calculate the value of the expression by applying the operations of the ring Z. Now similarly, we have expressions in the HTS. Here is one expression: ((Int -> Str) -> ([Int] -> [Str])) The above expression is fully parenthesized, and, thus we know what type it evaluates to. Here is another expression: (Int -> (Str -> ([Int] -> [Str]))) This expression too is fully parenthesized, and, thus we know what type it evaluates to. These two types specified by the two different expressions turn out not to be the same type. That is, the two expressions evaluate to different types. The concept of "currying" has come up. Part of my difficulty in learning a little bit of New Crazy Type Theory is that New Crazy Type Theorists have syntactic conventions which permit them to, often, reduce the number of parentheses in expressions. For example, one convention specifies that the string "a -> b -> c" should be translated to the fully parenthesized (a -> (b -> c)) and not to the different expression ((a -> b) -> c) The two expressions are different, and in general they evaluate to different types. And you are right: this convention has somewhat to do with currying. We say no more about currying, except that, to define "currying" a third operation on types is required: the product of types: xx: T x T -> T (Again, we write "xx" so that the operation on the left of ":" is not confused with the "x" on the right. Oi.) I speak now as a beginner. There is a phrase which, when I was an even more tenderly ignorant beginner than I am now, always confused me: Such and such operation is binary and right-associative. My delicate beginner's sensibilities were outraged. No! The operation might indeed be binary, but "right-associative" refers to the system of notation(s) for the operation(s), not to the operation. Of course, the above occasions for confusion are just the first and most simple, on the way to "hello world" in Haskell. I remain, as ever, your fellow student of history and probability, Jay Sulzberger PS. Likely there are errors in above.

without the parens (which is a natural result of lambda calculus, perhaps?) -- which is not what is meant by map. But underlying a Haskell type declaration is currying, is it not? At the type declaration level, it's all currying, correct?

Conceptually, I understand how the a -> b "event" needs to be a "package" to apply to the list [a]. The map function commandeers the target function (which alone by itself does some a -> b evaluation) to be a new object that is then applied to each member of list [a]. Good. So (a -> b) then is a notation that signifies this "package-ness".

Does anyone have examples of other "packaging" where a function doing some a -> b is changed to (a -> b) ?

On Fri, Dec 18, 2020 at 5:18 PM Bruno Barbier wrote:

...

Hi Lawrence,

Lawrence Bottorff writes:

...

Why is it not just

a -> b -> [a] -> [b]

again, why the parentheses?

In Haskell, (->) is a binary operator and is right associative. If you write:

a -> b -> [a] -> [b]

it implicitly means:

a -> (b -> ([a] -> [b]))

So here, you need explicit parenthesis:

(a -> b) -> [a] -> [b]

to mean: (a -> b) -> ([a] -> [b])

It's more about parsing binary operators than about types.

Does it help ?

Bruno

...

On Fri, Dec 18, 2020 at 4:10 PM Ut Primum wrote:

...

Hi,

a -> b is the type of a function taking arguments of a generic type (we call it a) and returning results of another type, that we call b.

So (a -> b ) -> [a] -> [b] Means that you have a first argument that is a function (a-> b), a second argument that is a list of elements of the same type of the function input, and that the returned element is a list of things of the type of the output of the function.

Cheers, Ut

Il ven 18 dic 2020, 23:02 Lawrence Bottorff ha scritto:

...

Thank you, but why in

map :: (a -> b) -> [a] -> [b]

are there parentheses around a -> b ? In general, what is the currying aspect of this?

On Fri, Dec 18, 2020 at 12:43 PM David McBride wrote:

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They are not parameters, they are the types of the parameters.

In this case a can really be anything, Int, Char, whatever, so long as the function takes a single argument of that type and the list that is given has elements of that same type. It is the same for b, it doesn't matter what b ends up being, so long as when you call that function the function's return value is compatible with the element type of the list that you intended to return from the entire statement.

You can mess with it yourself in ghci to see how type inference works.

> :t show :show :: Show a => a -> String > :t map show map show :: Show a => [a] -> [String] > :t flip map [1::Int] > flip map [1::Int] :: (Int -> b) -> [b]

On Fri, Dec 18, 2020 at 1:31 PM Lawrence Bottorff wrote:

> I'm looking at this > > ghci> :type map > map :: (a -> b) -> [a] -> [b] > > and wondering what the (a -> b) part is about. map takes a function > and applies it to an incoming list. Good. Understood. I'm guessing that the > whole Haskell type declaration idea is based on currying, and I do > understand how the (a -> b) part "takes" an incoming list, [a] and > produces the [b] output. Also, I don't understand a and b very well > either. Typically, a is just a generic variable, then b is another > generic variable not necessarily the same as a. But how are they being > used in this type declaration? > > LB > _______________________________________________ > Beginners mailing list > Beginners@haskell.org > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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