Why does Haskell's “do nothing” function, id, consume tons of memory?

问题

Haskell has an identity function which returns the input unchanged. The definition is simple:

id :: a -> a
id x = x

So for fun, this should output 8:

f = id id id id id id id id id id id id id id id id id id id id id id id id id id id
main = print $ f 8

After a few seconds (and about 2 gb of memory according to Task Manager), compilation fails with ghc: out of memory. Similarly, the interpreter says ghci: out of memory.

Since id is a pretty simple function, I wouldn't expect it to be a memory burden at run time or compile time. What is all the memory being used for?

回答1:

We know the type of id,

id :: a -> a

And when we specialize this for id id, the left copy of id has type:

id :: (a -> a) -> (a -> a)

And then when you specialize this again for the leftmost id in id id id, you get:

id :: ((a -> a) -> (a -> a)) -> ((a -> a) -> (a -> a))

So you see each id you add, the type signature of the leftmost id is twice as large.

Note that types are deleted during compilation, so this will only take up memory in GHC. It won't take up memory in your program.

来源：https://stackoverflow.com/questions/23746852/why-does-haskells-do-nothing-function-id-consume-tons-of-memory