[Haskell-beginners] Expected vs. Inferred Type

Daniel Fischer daniel.is.fischer at googlemail.com
Mon Jun 6 10:49:19 CEST 2011

On Montag, 6. Juni 2011, 05:07, Arlen Christian Mart Cuss wrote:
> On Sun, 2011-06-05 at 17:44 -0800, Christopher Howard wrote:
> > Could someone explain the difference between an expected type and an
> > inferred type as mentioned in a compiler error message? I've never
> > heard definitions proper, and so I tend to get confused when the
> > compiler says there is a difference between the type it expected and
> > the type it inferred.
> Hi Christopher,
> The expected type is the type expected by the compiler in that position;
> e.g. the argument of `not' should be a Bool:
> > :t not
> not :: Bool -> Bool
> Hence in the expression `not x', `x' is *expected* to be a Bool.
> The inferred (or actual) type is the type that Haskell has determined
> the value *actually* is.

Note however, that `expected' and `inferred' are not intrisic, which is 
which depends on the order of type-checking. For example:

Prelude> do { c <- "string"; putChar c; }

    Couldn't match expected type `[b]' against inferred type `IO ()'
    In the expression: putChar c
    In the expression:
        do c <- "string"
           putChar c
    In the definition of `it':
        it = do c <- "string"
                putChar c

so typechecking here proceeds left-to-right. If it proceeded right-to-left, 
you'd get an expected type of `IO a' and an inferred type of `[Char]'.

Also note that ghc-7.* says:

    Couldn't match expected type `[b0]' with actual type `IO ()'

i.e., uses `actual' instead of `inferred' in the hope that that is less 

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