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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