Hans Åberg wrote:
> >> Ah, your idea is to wrap std::move around the $k values! My idea
> >> is to let the C++ language recognise that, and then automate the
> >> assignment to $$.
> >
> > Again, the assignment to $$ is <bold>*NOT*</bold> the issue! (*)
> > It's the passing (moving) of $k.
> >
> > (*) Nitpick: Except in a case like "$$ = $2;", but then I'd argue
> > it's the passing of $2 to $$'s assignment operator that's the
> > issue. :)
>
> Maybe not for you, but for a more normal use, with types having
> both copy and move. Then one would like to use move to $$ whenever
> possible. Doesn't matter with me, as it is just some pointers and
> integers.
As I said, this already happens in "$$ = foo ();" (automatially) and
"$$ = std::move ($k);" (whether the move is explicit or
automatically inserted as we're discussing). So if $k is covered, no
special handling for $$ seems necessary.
> > You mean some kind of code analyzer? This might be possible, but may
> > be overkill.
>
> Yes, you might check how people do check against that problem, if
> there is some program doing that. Then one might get ideas of how
> to get into Bison.
>
> Perhaps it might be possible to have some DLL or external program and invoke
> that.
If you know of one, let me know. (And it would have to be
GPL-compatible in order to link it into Bison.)
> No, but it seems me it is a hard problem. A compiler optimizer can
> recognize such things if it has sufficient information about the
> types, by tracing the code flow.
Yes, perhaps we should ignore the issue for now and hope for
compilers to offer such a warning in the future (which would be more
useful anyway, since it would work for all code, not only Bison
grammars).
> > On the same token, Bison could then also (optionally) warn if some
> > $k which has a type is not used. I don't know if there is interest
> > in such a feature. It could also be useful for other languages.
> >
> > It might not be too hard to do for someone who's familiar with
> > Bison's internals. (Unfortunately, I'm not very much, and don't have
> > much free time right now.)
>
> How would this be different from the current static type system?
The type system wouldn't change. Just an additional check:
expr: expr '+' expr { $$ = $1; };
Obviously one forgot to use $3 here which has a type (unlike $2
which doesn't have a type and so is not expected to be used).
Bison could detect and warn about this (proably optional, since I
guess some people declare semantic types that are only meant to be
used sometimes).
Regards,
Frank
