Re: [Haskell-cafe] for = flip map

On 29 March 2012 22:03, Sjoerd Visscher wrote:

Some more bikeshedding:

Perhaps ffor, as in

   ffor = flip fmap

or perhaps

   infixr 0 <$$>    (<$$>) = flip (<$>)

   xs <$$> \x -> ...

I don't think it makes sense to add a whole new operator for that. You can just use sections: (<$> xs) \x -> ... The reason you can't do this with <*> is the ordering of effects. I have to admit, though, that the above isn't exactly readable. The non-operator version is somewhat more readable: (`map` xs) \x -> ... I'd still prefer "for" or "foreach".

(cf. <**>)

In both cases they should go in Data.Functor

Sjoerd

On Mar 28, 2012, at 11:26 PM, ezra@ezrakilty.net wrote:

...

I would very much like to see a standard function for "flip map" along these lines. I think it would make a lot of code more readable.

Like the OP, I use "for" in my own code. It's unfortunate that Data.Traversable takes the name with another type. Two options would be to (a) reuse the name in Data.List and force people to qualify as necessary, or (b) choose another name for "flip map".

Regarding other possible names: forall is a keyword and forAll is used by QuickCheck. One possibility would be "foreach".

Ezra

On Wed, Mar 28, 2012, at 10:19 PM, Christopher Done wrote:

...

On 28 March 2012 22:05, Matthew Steele wrote:

...

Doesn't for already exist, in Data.Traversable?   Except that for = flip traverse.

Traverse doesn't fit the type of fmap, it demands an extra type constructor:

traverse :: (Traversable t,Applicative f) => (a -> f b) -> t a -> f (t b)

fmap :: Functor f => (a -> b) -> f a -> f b

Note the (a -> f b) instead of (a -> b).

E.g.

fmap :: (a -> b) -> [a] -> [b]

can't be expressed with traverse, you can only get this far:

traverse :: (a -> [b]) -> [a] -> [[b]]

Unless I'm missing something.

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-- Sjoerd Visscher https://github.com/sjoerdvisscher/blog

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