Tony Morris wrote:

> is the same as:
> (.) :: (b -> c) -> ((a -> b) -> (a -> c))
> ..
> "accepts a function a to c and returns a function. The function returned
> takes a function a to b and returns a function a to c"
IMO, that should be
"accepts a function b to c and returns a function. The function returned
takes a function a to b and returns function a to c"
Personally I also find the following a good explanation, since it does not introduce lambdas or other scary things for newbies.

f . g = composite
    where composite x = f (g x)

for example

printLn = putStrLn . show

so

printLn x
   
becomes

putStrLn (show x)

Cheers,
Peter



 
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Apply parentheses from the right.

So:
(.) :: (b -> c) -> (a -> b) -> a -> c

is the same as:

(.) :: (b -> c) -> (a -> b) -> (a -> c)

is the same as:
(.) :: (b -> c) -> ((a -> b) -> (a -> c))

How you read that is up to you, but here is one way of reading it:

"accepts a function a to c and returns a function. The function returned
takes a function a to b and returns a function a to c"

The expression f(g(x)) in C-style languages is similar to (f . g) x

Tony Morris
http://tmorris.net/



PR Stanley wrote:
  
Hi
(.) :: (b -> c) -> (a -> b) -> (a -> c)
While I understand the purpose and the semantics of the (.) operator I'm
not sure about the above definition.
Is the definition interpreted sequentially - (.) is a fun taht takes a
fun of type (b -> c) and returns another fun of type (a -> b) etc?
Any ideas?
Thanks, Paul

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