Functional Purity using 'let' in Haskell

Question

As I am working on learning Haskell, I understand it is a purely functional language. I am having trouble understanding why let-statements don\'t violate purity

Answer 1

Your second let creates a new binding for e that shadows the existing variable. It does not modify e. You can easily check this with the following:

Prelude> let e = 1
Prelude> let f () = "e is now " ++ show e
Prelude> f ()
"e is now 1"
Prelude> let e = 2
Prelude> e
2
Prelude> f ()
"e is now 1"
Prelude>