如何在Haskell中实现列表推导？

Question

• Are list comprehensions simply a language feature?
• What's the easiest way to fake a list comprehension using pure Haskell?
• Do you have to use a do block/ >>= to do this or could you use some other method for hacking a list comprehension together?

Clarification: By "fake" a list comprehension I mean create a function that takes the same input and produces the same input, ie a form for the return values, lists to crunch together, and a predicate or multiple predicates.

Answer 1

Section 3.11 in the Haskell report describes exactly what list comprehensions mean, and how to translate them away.

If you want monad comprehensions you basically need to replace [e] by return e , [] by mzero , and concatMap by (>>=) in the translation.

Answer 2

To augment augustss 's answer, if you have something like:

[(x, y) | x <- [1..3], y <- [1..3], x + y == 4]

... it is equivalent to this use of do notation:

do x <- [1..3]
   y <- [1..3]
   guard (x + y == 4)
   return (x, y)

... which is equivalent to this use of concatMap :

concatMap (\x ->
    concatMap (\y ->
        if (x + y == 4) then [(x, y)] else []
        ) [1..3]
    ) [1..3]