Re: set $argv = *argv@100

Kevin Buettner
2002-02-12T06:22:51Z

On Feb 12, 12:47am, Andrew Cagney wrote:

> Consider the sequence:
>
> (top-gdb) print argv
> $4 = (char **) 0xbfbfd3cc
> (top-gdb) set $argv = *argv@100
> (top-gdb) print $argv
> $5 = (char **) 0xbfbfd3cc
> (top-gdb) print sizeof ($argv)
> $6 = 0x4
> (top-gdb) print sizeof (*argv@100)
> $7 = 0x190
>
> should $argv have been assigned the contents of *argv@100 or just the
> address?

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