Fw: [p4perl] sync problems

[Roger Day]

you are correct. You think you know a language, then it still manages to 
bite u in the foot. Doh!

Doesn't help me with my main problem in that the files still aren't being 
transferred to the new root.

Roger





Tony Smith <tony at smee.org>
14/03/2005 18:15
 
        To:     p4perl at perforce.com
        cc:     "Roger Day" <roger.day at globalgraphics.com>
        Subject:        Re: Fw: [p4perl] sync problems


On Monday 14 March 2005 17:09, Roger Day wrote:
> It's the same as before, except now:
>
> $#warnings == -1
>
> Roger

How's 'warnings' declared? $# operates directly on an array, not on a
reference to an array. i.e.

@warnings;
printf( "length = %d\n", $#warnings );

produces:

length = 0

But:

$warnings = $p4->Errors();
printf( "length = %d\n", $#warnings );

produces:

length = -1

Look familiar? In P4Perl it might look something like this:

#!/usr/bin/perl
use P4;

my $p4 = new P4;
my $errors = $p4->Errors();
my $warnings = $p4->Warnings();

printf( "%d (%d) errors\n", $#errors, $p4->ErrorCount() );
printf( "%d (%d) warnings\n", $#warnings, $p4->WarningCount() );

and this produces:

-1 (0) errors
-1 (0) warnings

This is because the $# operator is looking for @errors and @warnings, 
neither
of which have been defined. All brought to you through the joy of perl's
context-awareness....

If you make sure that @warnings and @errors are declared as arrays:

#!/usr/bin/perl
use P4;

my $p4 = new P4;
my @errors = $p4->Errors();
my @warnings = $p4->Warnings();

printf( "%d (%d) errors\n", $#errors, $p4->ErrorCount() );
printf( "%d (%d) warnings\n", $#warnings, $p4->WarningCount() );

You get:

0 (0) errors
0 (0) warnings

which is what you'd expect.

Tony

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