opStarAssign?

[0ffh]

Janice Caron wrote:
 > On 11/23/07, 0ffh <frank at frankhirsch.youknow.what.todo.net> wrote:
 >> Janice Caron wrote:
 >>> My preference would be to have the compiler rewrite
 >>>     *p += n;
 >>>
 >>> as
 >>>     p.opDerefAssign(p.opDeref() + n);
 >> Wouldn't "p.opAssign(p.opDeref()+n);" be quite sufficient?
 >
 > I think that would end up being equivalent to
 >     p = *p + n;

Si.

 > whereas what we actually want is the equivalent of
 >     *p = *p + n;

Is that so?
I'd guess that cases where both "x=y;" and "*x=y;" make sense are rare.
So overloading opAssign for the parameter type might be quite sufficient.

 > but I could be wrong. Maybe it depends on what we consider opAssign
 > ought to be doing. In any case, we /do/ have opIndexAssign(), so
 > opDerefAssign() ought to exist by analogy.

Well it's good that you know what ought to be - I'm sure I don't. ;-)

regards, frank