Ok but with these two record types defined :
    type t0 = { x : int ; y : int } ;;
    type t1 = { x : int } ;;

There is no ambiguity about the following expression's type:
    { x = 0 ; y = 0 } ;;
Why can't the t0 type be infered ?

And with this function definition,
    let get_y (t:t0) = t.x ;;
I explicitly say to the compiler that t is of type t0, so why does the 
compiler infer a t1 type when trying to get the x label of a t0 type value ?



Matthieur Dubuget wrote :
 > If there is y.x in your code, and the type t = {x : blabla; ...} was
 > declared,
 > then y's type is infered to be of type t

Skaller wrote :
 > So that the type of an expression:
 > { x = 1; y = 2 }
 > can be determined.