Question

I am new to perl and i have a question about perl thread.

I am trying to create a new thread to check if the running function is timed out, and my way of doing it is as below.

Logic is
1.create a new thread
2.run the main function and see if it is timed out, if ture, kill it

Sample code:

$exit_tread = false;      # a flag to make sure timeout thread will run
my $thr_timeout = threads->new( \&timeout );  
execute main function here;                

$exit_thread = true       # set the flag to true to force thread ends
$thr_timeout->join();      #wait for the timeout thread ends

Code of timeout function

sub timeout
{

$timeout = false;
my $start_time = time();
while (!$exit_thread)
{

    sleep(1);                 
    last if (main function is executed);

    if (time() - $start_time >= configured time )
    {
        logmsg "process is killed as request timed out";
        _kill_remote_process();
        $timeout = true;   
        last;         
    }
}    
}

now the code is running as i expected, but i am just not very clear if the code $exit_thread = true works because there is a “last” at the end of while loop.

Can anybody give me a answer?

Thanks

Answer 1

assuming that the correct pseudo code is this:

sub timeout
{

$timeout = false;
my $start_time = time();
#$timeout instead of $exit_thread 
while (!$timeout)
{

    sleep(1);                 
    last if (main function is executed);

    if (time() - $start_time >= configured time )
    {
        logmsg "process is killed as request timed out";
        _kill_remote_process();

        $timeout= true;   
        last;         
    }
}    
}

Then, no, calling last will not set the $timeout variable – it will simply leave the loop from the point of calling it.

Is this answering your question?