Variables in Haskell

Question

Why does the following Haskell script not work as expected?

find :: Eq a => a -> [(a,b)] -> [b]
find k t = [v | (k,v) <- t]

Giv

Answer 1

You can only pattern match on literals and constructors.
You can't match on variables. Read more here.

That being said, you may be interested in view patterns.

Answer 2

From the Haskell 98 Report:

As usual, bindings in list comprehensions can shadow those in outer scopes; for example:

[ x | x <- x, x <- x ] = [ z | y <- x, z <- y]

One other point: if you compile with -Wall (or specifically with -fwarn-name-shadowing) you'll get the following warning:

Warning: This binding for `k' shadows the existing binding
           bound at Shadowing.hs:4:5

Using -Wall is usually a good idea—it will often highlight what's going on in potentially confusing situations like this.

Answer 3

The pattern match (k,v) <- t in the first example creates two new local variables v and k that are populated with the contents of the tuple t. The pattern match doesn't compare the contents of t against the already existing variable k, it creates a new variable k (which hides the outer one).

Generally there is never any "variable substitution" happening in a pattern, any variable names in a pattern always create new local variables.