Haskell: trailing right arrow in a bind operation

Question

From haskellwiki: expr1 >>= \x ->

I am curious about the right side of the (>>=) operator \x ->. What does it mean?

Nor \x -> nor \x is recognized by :type in GHCi.

Answer 1

To be clear, the section you're quoting is this:

The actual translation from do notation to standard monadic operators is roughly that every expression matched to a pattern, x <- expr1, becomes

expr1 >>= \x ->

and every expression without a variable assignment, expr2 becomes

expr2 >>= \_ ->

All do blocks must end with a monadic expression, and a let clause is allowed at the beginning of a do block (but let clauses in do blocks do not use the in keyword). The definition of mothersPaternalGrandfather above would be translated to:

mothersPaternalGrandfather s = mother s >>= \m ->
                           father m >>= \gf ->
                           father gf

So as you can see, the -> is not actually trailing. If you look at the final example in the quotation above, where mothersPaternalGrandfather is defined, you'll see that the ->s all have a right hand side, which continues on the next line. The last line in the definition doesn't end with ->.

As the text on the wiki explains, the expr1 >>= \x -> is just "roughly" what happens. You're right that expr1 >>= \x -> is not valid syntax. A more fully parenthesized version of the function definition would look like this:

mothersPaternalGrandfather s =
  mother s >>= (\m -> father m >>= (\gf -> father gf))