Hi, I can understand why principle type of map is map :: (a -> b) -> [a] -> [b] , I would interpret this as "map takes a function of type a->b and a list of type [a] as arguments and returns a list of type [b]" but it is going somewhat beyond my imagination why principle type of map map is (map map)::[a -> b] -> [[a] -> [b]] I am able to interpret the expressions "[a -> b] -> [[a] -> [b]]" vaguely... does this mean that 'map map' takes list of functions of type (a->b) and returns list of functions of type ([a]->[b]) if yes ..how do I derive it from basic type declaration of map? Thanks in advance Vikrant
Vikrant wrote:
(map map)::[a -> b] -> [[a] -> [b]]
I am able to interpret the expressions "[a -> b] -> [[a] -> [b]]" vaguely...
does this mean that 'map map' takes list of functions of type (a->b) and returns list of functions of type ([a]->[b]) if yes ..how do I derive it from basic type declaration of map?
Think substitution. It helps if you distinguish the letters. Start by saying that one of the maps has type (c -> d) -> [c] -> [d]. Then substitute a =(c->d), b=[c] (I think it would be). Paul.
Vikrant wrote:
Hi, I can understand why principle type of map is
map :: (a -> b) -> [a] -> [b] , I would interpret this as "map takes a function of type a->b and a list of type [a] as arguments and returns a list of type [b]"
but it is going somewhat beyond my imagination why principle type of map map is
(map map)::[a -> b] -> [[a] -> [b]]
I am able to interpret the expressions "[a -> b] -> [[a] -> [b]]" vaguely...
does this mean that 'map map' takes list of functions of type (a->b) and returns list of functions of type ([a]->[b]) if yes ..how do I derive it from basic type declaration of map?
Paul's hint is more or less the same as this, but: map's first parameter has type (a->b). You are passing it 'map' as the first parameter. 'map' has type (a->b) -> ([a]->[b]), so what are my first 'a' and 'b' in this case? (renaming may make it clearer). If you get that cleared up, you may be interested in the type of (map . map) and why that happens :) Jules
Hi Vikrant, The other two have said basically the same thing as me, but your description of map's type makes me think maybe an explanation of curried functions is in order: map is what's called a curried function. Basically, the type: (a -> b) -> [a] -> [b] has two interpretations. The first is the one you gave, which is a good way to think about the function when you intend to apply it to both its arguments at once. However, there is another way to think about the function as well. Consider that the type implicitly has parenthesis like this: (a -> b) -> ([a] -> [b]) That is, we can also think of map as a function which takes a function from a to b, and returns a function from [a] to [b]. In some sense, if we only provide map with its first argument, then it is still "waiting" to take the [a] argument, so it becomes a function from [a] to [b]. If we think of may this way, we can see that when it is passed to itself, its type must be matched against the type of its first argument. I'll rename the type variables to make things a little clearer - what we have to do is unify: (c -> d) -> ([c] -> [d]) and a -> b So in this case, when map returns a function from [a] -> [b], you get a function from [c -> d] to [[c] -> [d]]. I hope this helps! --Chris Casinghino On Wed, 14 Mar 2007, Vikrant wrote:
Hi, I can understand why principle type of map is
map :: (a -> b) -> [a] -> [b] , I would interpret this as "map takes a function of type a->b and a list of type [a] as arguments and returns a list of type [b]"
but it is going somewhat beyond my imagination why principle type of map map is
(map map)::[a -> b] -> [[a] -> [b]]
I am able to interpret the expressions "[a -> b] -> [[a] -> [b]]" vaguely...
does this mean that 'map map' takes list of functions of type (a->b) and returns list of functions of type ([a]->[b]) if yes ..how do I derive it from basic type declaration of map?
Thanks in advance Vikrant
woow :) I am loving this language.
Thank you all for very useful hints and thanks for the exercise about
map.map!
On 14/03/07, Big Chris
Hi Vikrant,
The other two have said basically the same thing as me, but your description of map's type makes me think maybe an explanation of curried functions is in order:
map is what's called a curried function. Basically, the type:
(a -> b) -> [a] -> [b]
has two interpretations. The first is the one you gave, which is a good way to think about the function when you intend to apply it to both its arguments at once. However, there is another way to think about the function as well. Consider that the type implicitly has parenthesis like this:
(a -> b) -> ([a] -> [b])
That is, we can also think of map as a function which takes a function from a to b, and returns a function from [a] to [b]. In some sense, if we only provide map with its first argument, then it is still "waiting" to take the [a] argument, so it becomes a function from [a] to [b].
If we think of may this way, we can see that when it is passed to itself, its type must be matched against the type of its first argument. I'll rename the type variables to make things a little clearer - what we have to do is unify:
(c -> d) -> ([c] -> [d]) and a -> b
So in this case, when map returns a function from [a] -> [b], you get a function from [c -> d] to [[c] -> [d]]. I hope this helps!
--Chris Casinghino
On Wed, 14 Mar 2007, Vikrant wrote:
Hi, I can understand why principle type of map is
map :: (a -> b) -> [a] -> [b] , I would interpret this as "map takes a function of type a->b and a list of type [a] as arguments and returns a list of type [b]"
but it is going somewhat beyond my imagination why principle type of map map is
(map map)::[a -> b] -> [[a] -> [b]]
I am able to interpret the expressions "[a -> b] -> [[a] -> [b]]" vaguely...
does this mean that 'map map' takes list of functions of type (a->b) and returns list of functions of type ([a]->[b]) if yes ..how do I derive it from basic type declaration of map?
Thanks in advance Vikrant
participants (4)
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Big Chris -
Jules Bean -
Paul Johnson -
Vikrant