Re^4: logfile rotate w/ pass by reference

ok so in your sample code how does the subroutine know what variable it is receiving based off of your filename? Im ny code I have two distinct filenames I want to pass to logfile::rotate so how do I do this?


        if ( ( my $dif = $dif[1] - $dif[2] ) > 199 ) {
                &mailme;
                logroll($out);
       
        if ( ( my $diff = $diff[1] - $diff[2] ) > 199 ) {
                &mailme;
                logroll($out1);

        sub logroll {
                my $logs = new Logfile::Rotate 
                (File => ????  # <---- what is here based?
                Count => 10,   # off of 2 diff filenames
                Gzip => 'lib',
                Dir  => '/usr/local/log/old',
                Flock => 'yes',
                Persist => 'yes' );
                $logs->rotate();
        }
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Comment on Re^4: logfile rotate w/ pass by reference Download Code

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Re^5: logfile rotate w/ pass by reference by Joost (Canon) on Mar 14, 2005 at 17:53 UTC
It doesn't need to know the name of the variable that is being passed. That's whole point of having subroutine arguments in the first place. All your logroll function needs to know is the name of the file. The name of the file is the value of the argument. The name of the variable is not important. `sub print_value { my $value = shift; print $value; } my $var1 = "value in var1\n"; my $var2 = "value in var2\n"; print_value($var1); print_value($var2);` [download] This is all documented in the perlfunc manpage. update: I see I mistyped the code in the previous example. Maybe my update will make it clearer. :-) "What should it profit a man, if he should win a flame war, yet lose his cool?"	[reply] [d/l]

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Re^5: logfile rotate w/ pass by reference
by Joost (Canon) on Mar 14, 2005 at 17:53 UTC

All your logroll function needs to know is the name of the file. The name of the file is the value of the argument. The name of the variable is not important.

sub print_value {
  my $value = shift;
  print $value;
}

my $var1 = "value in var1\n";
my $var2 = "value in var2\n";

print_value($var1);
print_value($var2);
[download]

This is all documented in the perlfunc manpage.

update: I see I mistyped the code in the previous example. Maybe my update will make it clearer. :-)

"What should it profit a man, if he should win a flame war, yet lose his cool?"

[reply]
[d/l]