Re: Is cast operator unary or binary? How many operands?

[JoseMariaSola]

On Apr 29, 1:47 pm, JoseMariaSola <JoseMariaS...@xxxxxxxxx> wrote:
> > > I'd say one operator, the cast operator, and two operands: typename
> > > and expression.
>
> > > But everywhere I read, cast is categorized as an unary operator. Why
> > > is that? Is it just a syntax cotegory?
>
> > (typename)(expression) has one operator `(typename)' and one operand
> > `(expression)'.
> > The reason the type is enclosed in parentheses (as a design decision)
> > is probably to avoid ambiguity, consider this:
>
> > int i = 1; /* define and initialize i to 1 */
> > {
> >   int i; /* cast i to int, a statement with no effect, or define i in
> > block scope? */
>
> > }
>
> Thanks, Vipps.
>
> According to your answeer, the operator '(typename)' is very
> particular, because it not a single token but three AND the middle
> token is anything an identifier may be.
>
> JM.

The last line of my last post shoudl be:
... the middle token is anything an identifier may be and more.

Here is part of the grammar:

unary-expression:
    postfix-expression
    ++ unary-expression
    -- unary-expression
    unary-operator cast-expression
    sizeof unary-expression
    sizeof ( type-name )

unary-operator: one of
    & * + - ˜ !

cast-expression:
    unary-expression
    ( type-name ) cast-expression


Why sizeof, (type-name), ++ and -- aren't unary-operators?

Thanks.