Type checker's expected and inferred types (reformatted)
dave at zednenem.com
Sun Oct 25 16:02:51 EDT 2009
On Sun, Oct 25, 2009 at 1:37 PM, Isaac Dupree
<ml at isaac.cedarswampstudios.org> wrote:
> David Menendez wrote:
>> The expected type is what the context wants (it's *ex*ternal). The
>> inferred type is what the expression itself has (it's *in*ternal).
>> So inferring the type Maybe () for bar seems wrong.
> well, maybe GHC just gets it wrong enough of the time, that I got confused.
> Or maybe ... When there are bound variables interacting, on the inside and
> outside, it gets confusing.
> Prelude> \x -> (3+x) + (length x)
> Couldn't match expected type `[a]' against inferred type `Int'
> In the second argument of `(+)', namely `(length x)'
> In the expression: (3 + x) + (length x)
> In the expression: \ x -> (3 + x) + (length x)
> Your explanation of "expected" and "inferred" could make sense to me if the
> error message followed the "Couldn't match" line with, instead,
> "In the first argument of `length', namely `x'"
> because 'length' gives the context of expected list-type, but we've found
> out from elsewhere (a vague word) that 'x' needs to have type Int.
This had me confused for a while, but I think I've worked out what's
happening. (+) is polymorphic, and GHC is giving it the type [a] ->
[a] -> [a]. So the context is expecting [a], but we infer length x ::
Int from the definition of length.
In the alternate case, \x -> length x + (3+x), GHC gives the outer (+)
the type Int -> Int -> Int, and the inner (+) the type [a] -> [a] ->
[a], which is why we get the type mismatch complaint for 3+x instead
Note what happens if we use a monomorphic operator:
Prelude> let (<>) = undefined :: Int -> Int -> Int
Prelude> \x -> (3+x) <> length x
Couldn't match expected type `[a]' against inferred type `Int'
In the first argument of `length', namely `x'
In the second argument of `(<>)', namely `length x'
In the expression: (3 + x) <> length x
Prelude> \x -> (3+x) + length x
Here, GHC has concluded that x must be an Int, and thus can't be
passed to length.
Prelude> \x -> length x <> (3+x)
Couldn't match expected type `Int' against inferred type `[a]'
In the second argument of `(<>)', namely `(3 + x)'
In the expression: length x <> (3 + x)
In the expression: \ x -> length x <> (3 + x)
Here, GHC has concluded that x must be [a], and thus 3+x must be [a],
which can't be used with <>.
Dave Menendez <dave at zednenem.com>
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