Perl parameters are passed by reference - see perlsub. However, when you do
my ($param1, $param2) = @_$param1 and $param2 get a copy of each element in @_, effectively turning a pass by reference into a pass by value. If you need to access the original variable passed in rather than a copy of its value, you can use $_[0], $_[1], etc.
But I don't think passing by reference vs. value is much of an issue in the code sample above. Perl "objects" are references in their own right. If you pass $q to your function, then $queue and $q both point to the same object, not a copy of each other.
Additionally, new isn't a key word in Perl. It is a method that must be defined for each class. new Thread::Queue is simply a different spelling for Thread::Queue->new(). There is nothing magical about the name "new" that turns it into a constructor. It acts like a constructor because, by convention, this method returns a blessed reference to some data that it declared and initialized. For more information on Perl's approach to objects, please see perltoot and perlobj. You may also want to take a look at perlref for a better understanding of references in Perl.
All the above applies to function calls and variables within a thread. In Perl, all variables are, by default, thread local (see thread). Passing references between threads will cause thread death unless the data is explicitly marked as shared - see threadtut and threads::shared for more information.
Best, beth
In reply to Re: creating thread : are parameter objects passed via reference ?
by ELISHEVA
in thread creating thread : are parameter objects passed via reference ?
by cbrauner
| For: | Use: | ||
| & | & | ||
| < | < | ||
| > | > | ||
| [ | [ | ||
| ] | ] |