Haskell Generator with variable number of inputs?

Question

at the moment my code looks a bit like this:

gen (x:[]) = [[a] | a <- (someOp x)]
gen (x:y:[]) = [[a,b] | a <- (someOp x), b <- (someOp y)]
gen (x:y:z:[]) = [[a,b,c] | a <- (someOp x), b <- (someOp y), c <- (someOp z)]

... and so on

is it possible to conclude the rest with gen(x:xs) ??

Answer 1

You can do this recursively:

gen [] = [[]]
gen (x:xs) = [ a:g | a <- someOp x, g <- gen xs ]

At each step you take all the lists generated by the previous step and combine each of them with every result of someOp.

You can verify that this degrades into your special cases by substitution.

Answer 2

For the function you here construct, a more generic function already exists: traverse :: (Applicative f, Traversable t) => (a -> f b) -> t a -> f (t b). Your gen is equivalent to:

gen = traverse someOp