[Haskell-begin] some basic syntax questions
Niels Aan de Brugh
nielsadb at gmail.com
Fri Jul 25 16:26:16 EDT 2008
On Fri, 2008-07-25 at 22:45 +0300, Anatoly Vorobey wrote:
> 2. I initially tried
>
> where sumIt = sum $ map read
>
> (the "point-free" style, I believe it's called?) but that didn't
> compile. Why not? A friend suggested
>
> where sumIt = sum . map read
>
> and that does work; I guess my real problem, then, is that I don't
> really understand the difference
> between the two and why the former doesn't work.
>
The $ is an operator that performs function application. So for example:
show $ "Hello"
is equivalent to
show "Hello"
You don't need an operator for function application, normally you just
type your arguments after the function. But $ just happens to be
right-associative, which means it will "evaluate" (modulo lazy semantics
of course) its right-hand-side argument before applying it to its
left-hand-side argument. So your expression
sum $ map read
First "evaluates" (map read) then applies it to sum. That is, it's the
same as:
sum (map read)
This is a type error, as the compiler will tell you.
The function composition operator (.) takes two functions f :: (b->c)
and g :: (a->b) and returns a function (a->c) that effectively first
applies g (yielding a 'b') and then f (yielding a 'c'). That is: f after
g. So:
sum . map read
which should be read as:
sum . (map read)
which is equivalent to:
(\x -> sum ((map read) x))
which is itself a new function that reads a list of strings and sums the
result.
In my opinion it's often helpful to look at the function types using :t
in ghci, e.g. ":t (.)" and ":t ($)" (don't forget the brackets around
the operators).
I'm not an experienced Haskell programmer myself but I'd suggest not
using point free style if you find it confusing, but rather write out
parameter names and lambdas explicitly. You style will change as you get
more comfortable with the language.
Regards,
Niels
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