fis@wiwi.hu-berlin.de wrote:
> 
>>  let good : (int * float) = let x                  = fst in (x (0,1), x (0., 1.));;
>>  let bad  : (int * float) = let x: ('a * 'a) -> 'a = fst in (x (0,1), x (0., 1.));;
> 
> The type checker sais:
> 
>>  let bad  : (int * float) = let x: ('a * 'a) -> 'a = fst in (x (0,1), x (0., 1.));;
>>                                                                         ^^^^^^^^
>>This expression has type float * float but is here used with type int * int

This is due to OCaml's counter-intuitive interpretetion of type 
variables in type annotations. The 'a is not polymorphic! Rather, it is 
considered a monomorphic placeholder for some concrete type (IOW it is 
not quantified). The effect is similar to saying

   let x: (_ * _) -> _ = fst in (x (0,1), x (0., 1.))

except that you put the additional constraint that all occurances of the 
wildcard have to denote the same type.

When the type checker sees the first application of x it then knows that 
that 'a has to denote int. So it behaves as if you had written

   let x: (int * int) -> int = fst in (x (0,1), x (0., 1.))

(Personally, I never understood the rationale for that treatment of type 
variables. It is one of the things I find more reasonable in SML, where 
the equivalent of your example would type-check.)

Cheers,

     - Andreas

-- 
Andreas Rossberg, rossberg@ps.uni-sb.de

Let's get rid of those possible thingies!  -- TB

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