Why are (constant) expressions not evaluated at compile time in Haskell?

问题

I am currently learning Haskell, and there is one thing that baffles me:

When I build a complex expression (whose computation will take some time) and this expression is constant (meaning it is build only of known, hard coded values), the expression is not evaluated at compile time.

Comming from a C/C++ background I am used to such kind of optimization.

What is the reason to NOT perform such optimization (by default) in Haskell / GHC ? What are the advantages, if any?

data Tree a =
   EmptyTree
 | Node a (Tree a) (Tree a)
 deriving (Show, Read, Eq)

elementToTree :: a -> Tree a
elementToTree x = Node x EmptyTree EmptyTree

treeInsert :: (Ord a) => a -> Tree a -> Tree a
treeInsert x EmptyTree = elementToTree x
treeInsert x (Node a left right)
  | x == a = Node x left right
  | x < a  = Node a (treeInsert x left) right
  | x > a  = Node a left (treeInsert x right)

treeFromList :: (Ord a) => [a] -> Tree a
treeFromList []     = EmptyTree
treeFromList (x:xs) = treeInsert x (treeFromList xs)

treeElem :: (Ord a) => a -> Tree a -> Bool
treeElem x EmptyTree = False
treeElem x (Node a left right)
  | x == a = True
  | x < a  = treeElem x left
  | x > a  = treeElem x right

main = do
  let tree = treeFromList [0..90000]
  putStrLn $ show (treeElem 3 tree)

As this will always print True I would expect the compiled programm to print and exit almost immediately.

回答1:

You may like this reddit thread. The compiler could try to do this, but it could be dangerous, as constants of any type can do funny things like loop. There are at least two solutions: one is supercompilation, not available as part of any compiler yet but you can try prototypes from various researchers; the more practical one is to use Template Haskell, which is GHC's mechanism for letting the programmer ask for some code to be run at compile time.

回答2:

The process you are talking about is called supercompilation and it's more difficult than you make it out to be. It is actually one of the active research topics in computing science! There are some people that are trying to create such a supercompiler for Haskell (probably based on GHC, my memory is vague) but the feature is not included in GHC (yet) because the maintainers want to keep compilation times down. You mention C++ as a language that does this – C++ also happens to have notoriously bad compilation times!

Your alternative for Haskell is to do this optimisation manually with Template Haskell, which is Haskells compile-time evaluated macro system.

回答3:

In this case, GHC can not be sure that the computation would finish. It's not a question of lazy versus strict, but rather the halting problem. To you, it looks quite simple to say that treeFromlist [0..90000] is a constant that can be evaluated at compile time, but how does the compiler know this? The compiler can easily optimize [0..90000] to a constant, but you wouldn't even notice this change.

来源：https://stackoverflow.com/questions/19259114/why-are-constant-expressions-not-evaluated-at-compile-time-in-haskell