[Haskell-beginners] map type explanation
Jay Sulzberger
jays at panix.com
Sat Dec 19 05:44:32 UTC 2020
On Fri, 18 Dec 2020, Lawrence Bottorff <borgauf at gmail.com> 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 <brubar.cs at gmail.com> wrote:
>
>>
>> Hi Lawrence,
>>
>> Lawrence Bottorff <borgauf at gmail.com> 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 <utprimum at gmail.com> 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 <borgauf at gmail.com> 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 <toad3k at gmail.com>
>> wrote:
>>>>>
>>>>>> 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 <borgauf at gmail.com>
>>>>>> 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
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>>>>>>>
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