Re: set $argv = *argv@100

Andrew Cagney
2002-02-12T15:32:17Z

>
> 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 :-(

sigh,
Andrew