Pattern matching variables in a case statement in Haskell

Question

If I compare a string literal to a string literal using the case statement, I get the expected behavior: if they are the same - it matches, if they are not - it does not.

However, if I compare a string literal to a constant that is a string, I get "Pattern matches are overlapped" warning and the branch with the constant always matches.

Here's an example session:

Prelude> let var1 = "abc"
Prelude> let var2 = "def"
Prelude> case var1 of { var2 -> "Fail"; _ -> "Win" }

<interactive>:1:0:
    Warning: Pattern match(es) are overlapped
             In a case alternative: _ -> ...
"Fail"
Prelude> case "abc" of { var2 -> "Fail"; _ -> "Win" }

<interactive>:1:0:
    Warning: Pattern match(es) are overlapped
             In a case alternative: _ -> ...
"Fail"
Prelude> case "abc" of { "def" -> "Fail"; _ -> "Win" }
"Win"

Meanwhile, if behaves as expected:

> Prelude> if var1 == var2 then "Fail" else "Win" 
"Win"

What's going on here? How does this behavior make sense?

Answer 1

See Don's answer for why. A common idiom for doing what you are trying to do is this:

var1 = "abc"
var2 = "def"

foo x = case () of
    () | x == var1 -> "Fail"
       | x == var2 -> "Failzor"
       | otherwise -> "WIN"

Of course in this case we would lose the case and just write the guards directly on the function:

foo x | x == var1 = "Fail"
      | ...

UPDATE

These days the MultiWayIf extension does this with slightly less syntactic noise.

{-# LANGUAGE MultiWayIf #-}

foo x = if | x == var1 -> "Fail"
           | x == var2 -> "Failzor"
           | otherwise -> "WIN"