set $argv = *argv@100

set $argv = *argv@100

[Andrew Cagney]

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?

Contrast it with:

(top-gdb) print args
$9 = {func = 0x804d498 <captured_main>, func_args = 0xbfbfd8d4}
(top-gdb) set $a = args
(top-gdb) print $a
$10 = {func = 0x804d498 <captured_main>, func_args = 0xbfbfd8d4}

puzzled.

Andrew

Re: set $argv = *argv@100

[Kevin Buettner]

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

Re: set $argv = *argv@100

[Andrew Cagney]

> 
> 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

Re: set $argv = *argv@100

[Kevin Buettner]

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