Question

I’m just getting started on monads, and I can’t figure out why these two expressions evaluate differently:

ghci> [1,2] >>= \n -> ['a','b'] >>= \ch -> return (n,ch)
[(1,'a'),(1,'b'),(2,'a'),(2,'b')]


ghci> return ([1,2],['a','b'])
([1,2],"ab")

Answer 1

The types are different, so it’s reasonable that the behaviors are different

The first expression would typecheck as Num t => [(t, Char)]

The use of [] as the monad in the (>>=) means that it infers that the monad should be the List monad and in the context of the List Monad http://en.wikibooks.org/wiki/Haskell/Understanding_monads/List (>>=) is concatMap and return is (:[]).

[1,2] >>= \n -> ['a','b'] >>= \ch -> return (n,ch)

is the same as

concatMap (\n -> concatMap (\ch -> [(n, ch)]) ['a', 'b']) [1,2]

which returns [(1,'a'),(1,'b'),(2,'a'),(2,'b')]

In your second example what’s really going on is

that the type of the expression is a bit more general:

Prelude> :t return ([1,2],['a','b'])
return ([1,2],['a','b']) :: (Monad m, Num t) => m ([t], [Char])

Because you’re running it in GHCi a few things happen. GHCi can be considered a very big special IO Monad. So since no monad was specified, when GHC tries to print the results, it’ll take the m Monad to be IO in this case.

The t is defaulted to Integer as well, so the type of the resulting expression is :: IO ([Integer], [Char]).

As it happens, all the types used have a Show instance, so GHC can print the results of executing the IO action, which in this case (due to the action being return) is the same as the input.