Re: how cast works?

Ben Bacarisse <ben@bsb.me.uk> writes:

Keith Thompson <Keith.S.Thompson+u@gmail.com> writes:
>

Tim Rentsch <tr.17687@z991.linuxsc.com> writes:
[...]
>

Furthermore, there are cases where having to do a conversion from
one type to the same type has semantic consequences, even though
the types are the same.

>
What are these cases?

>
The case that comes to mind is where the result of a floating point
expression is represented with more precision than the type can hold
when stored.

Right.

6.3.1.8 p2:
  The values of floating operands and of the results of floating
  expressions may be represented in greater range and precision than
  that required by the type;  the types are not changed thereby.[63]
>
The footnote explains that the implicit conversion done on assignment
will change the value even though the types may be the same:
>
[63] The cast and assignment operators are still required to remove
     extra range and precision.

A related item is that the rule for 'return' is not the same as
assignment.  When returning a value from a function, the value
of the return expression is converted to the return type of the
function, /unless/ the type of the return expression is the same
as the return type of the function, in which case no conversion
takes place (and so extra range and precision may be retained).