Re: set $argv = *argv@100

Kevin Buettner
2002-02-12T16:33:07Z

On Feb 12, 10:32am, Andrew Cagney wrote:

> > I think the address is right. The problem is that $argv isn't the right
> > type. According to the above, it's ``char **'' when it should be
> > ``char *[100]''.
> >
> > Try this:
> >
> > (top-gdb) set $argva = &(*argv@100)
> > (top-gdb) print $argva
> > $23 = (char *(*)[100]) 0xbffff99c
> > (top-gdb) ptype $argva
> > type = char *(*)[100]
> > (top-gdb) ptype *$argva
> > type = char *[100]
> >
> > Kevin
>
> Yes, C strikes again. You can't copy an array, just its address :-(

I still think it'd be nice if we could somehow preserve the type
when doing set $argv = *argv@100...

(top-gdb) set $argv = *argv@100
(top-gdb) ptype $argv
type = char **
(top-gdb) ptype *argv@100
type = char *[100]

I.e, it'd be nice if the types of $argv and *argv@100 were the same.

I haven't thought this through though. It could be that a lot of other
stuff would break if we did this.

Kevin