Haskell: Do we have short cut for mapCons

I mean write such a funct开发者_高级运维ion：

mapCons x [] = [[x]]
mapCons x ys = map (x:) ys

so that:

*> mapCons 'a' []
["a"]
*> mapCons 'a' ["cat", "dog"]
["acat","adog"]

Do we have short-cut or pre-defined functions for mapCons? I tried Hoogle a->[[a]]->[[a]], but didn't find ideal one.

Thanks a lot.

No, and I note that map (x:) is 8 characters long while typing mapCons x is 9 characters. For such specific operations (i.e. non-generic operations) the cost and difficulty of finding or remembering them far out weighs the benefits of having them. Imagine if the prelude or base libraries defined 100,000 functions - shudder!

If you import Control.Applicative you can do things like this:

GOA Control.Applicative> (:) 'a' <$> ["cat", "dog"]
["acat","adog"]
GOA Control.Applicative> (:) <$> [0] <*> [[1,2,3], [7,8,9]]
[[0,1,2,3],[0,7,8,9]]
GOA Control.Applicative> liftA2 (:) "a" ["cat", "dog"]
["acat","adog"]

I know this doesn't satisfy the a->[[a]]->[[a]] signature you posted, but your question is a bit unclear as to what you are trying to do so I thought this still might help.

You can easily replace mapCons by a simple list comprehension:

mapCons x ys = [x:y | y<-ys]

Like the other solutions, it returns [] if fed with empty ys, not [[x]] as your definition.

If we disregard the empty-list case, which, as FUZxxl noticed, does not make much sense, you may notice that:

mapCons = map . (:)

Which, when read out loud, yields something tautological like "mapCons is equal to map composed with cons".

Does not seem useful to me.