Type inference interferes with referential transparency

Question

What is the precise promise/guarantee the Haskell language provides with respect to referential transparency? At least the Haskell report does not mention this notion.

C

Answer 1

What you think this has to do with referential transparency? Your uses of 7, ^, mod, 5, 2, and == are applications of those variables to dictionaries, yes, but I don't see why you think that fact makes Haskell referentially opaque. Often applying the same function to different arguments produces different results, after all!

Referential transparency has to do with this expression:

let x :: Int = 7^7^7`mod`5`mod`2 in (x == 1, [False, True] !! x)

x is here a single value, and should always have that same single value.

By contrast, if you say:

let x :: forall a. Num a => a; x = 7^7^7`mod`5`mod`2 in (x == 1, [False, True] !! x)

(or use the expression inline, which is equivalent), x is now a function, and can return different values depending on the Num argument you supply to it. You might as well complain that let f = (+1) in map f [1, 2, 3] is [2, 3, 4], but let f = (+3) in map f [1, 2, 3] is [4, 5, 6] and then say "Haskell gives different values for map f [1, 2, 3] depending on the context so it's referentially opaque"!