Does Haskell have variables?

Question

I\'ve frequently heard claims that Haskell doesn\'t have variables; in particular, this answer claims that it doesn\'t, and it was upvoted at least nine times and accepted.<

Answer 1

Yes, Haskell has variables. Consider the (essentially equivalent) definitions

inc n = n + 1
inc = \n -> n + 1

In both these cases, n is a variable; it will take on different values at different times. The Haskell Report, in Section 3 refers to these explicitly as variables.

That n here is a variable may be easier to see if we consider the following complete program:

inc n = n + 1
f = inc 0
g = inc 1
main = print (f+g)

The answer printed will be "3", of course. When evaluating f, as we expand inc x will take on the value 0, and when later (or earlier!) evaluating g, as we expand inc x will take on the value 1.

Some confusion may have arisen because Haskell, as with the other languages listed in the question, is a single-assignment language: it does not allow the reassignment of variables within a scope. Once n has been assigned the value 42, it cannot be anything but 42 without introducing a new scope with a new n (which is a different variable, shadowing the other n) bound to another value.

This may not be entirely obvious in some contexts, such as expressions using do:

 do let n = 1
    print n
    let n = 2
    print n

but if you remove the syntactic sugar, translating it into Haskell without the do, it becomes clear that there was a new, nested scope created where the n in that inner scope is a different variable that is shadowing the n in the outer scope:

(let n = 1 
  in (print n >> (let n = 2
                   in print n)))