Jon Harrop wrote:

>Considering the following code objects to the expression in the "for" loop not 
>having the type "unit":
>
># for i=0 to 3 do 1 done;;
>Warning: this expression should have type unit.
>- : unit = ()
>
>why is "g" in the following code inferred to have the polymorphic type (unit 
>-> 'a) rather than (unit -> unit)?
>
># let f g = for i=0 to 3 do g () done;;
>val f : (unit -> 'a) -> unit = <fun>
>  
>
Consider the fact that it's a warning as opposed to an error.  An 
expression with a result that isn't used (fairly rare in OCaml) is a 
special case; any type is permitted, but an explicitly non-unit type 
generates a warning, since it's probably an error.

If the type system were more advanced, your second example would infer a 
type of "unit -> 'a <warning-if-non-unit>".

If your second example produced a type of unit -> unit for g, I think 
your first example should fail with an error (rather than be accepted 
with a warning).

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