integer division with float result

[0ffh]

0ffh wrote:
> With return type overloading I get
>   [...]
>   foo(cast(int)i/j); // explicit int
>   foo(cast(float)i/j); // explicit float

Oops; this is with syntactic sugar, BTW, as I was
just casting the first operand, not the expression.
Otherwise
   foo(cast(int)(i/j)); // explicit int
   foo(cast(float)(i/j)); // explicit float

(The current style example should work as it is.)

Regards, Frank