Reputation: 11
I am working with ruby 192 p290: under one unit-test script (shown below) throws ThreadError
1) Error:
test_orchpr_pass(TC_MyTest):
ThreadError: deadlock; recursive locking
internal:prelude:8:in `lock'
internal:prelude:8:in `synchronize'
testth.rb:121:in `orchpr_run'
testth.rb:158:in `test_orchpr_pass'
With ruby 187 gives error: Thread tried to join itself.
def orchpr_run(timeout = 60)
# used by the update function to signal that a final update was
# received from all clients
@update_mutex.lock
# required since we'll have to act as an observer to the DRb server
DRb.start_service
# get configuration objects
run_config_type = DataLayer.get_run_config_type
client_daemon = DataLayer.get_client_daemon_by_branch (run_config_type, @branch)
client_daemon['port_no'] = 9096
#get the servers for this client_daemon
servers = DataLayer.get_servers(run_config_type, client_daemon.id)
servers.each { |server| @pr[server.host_name] = OrchestratedPlatformRun.new(run_config_type, server, timeout)
}
@pr.each_value { |x| x.add_observer(self)
@pr.each_value { |x| x.start(@service_command_pass, true)
# wait for update to receive notifications from all servers # this is the statement causing error:
@update_mutex.synchronize {} end
Another piece of code throwing same error:
require "thread"
require "timeout"
def calc_fib(n)
if n == 0
0
elsif n == 1
1
else
calc_fib(n-1) + calc_fib(n-2)
end
end
lock = Mutex.new
threads = 20.times.collect do
Thread.new do
20.times do
begin
Timeout.timeout(0.25) do
lock.synchronize{ calc_fib(1000) }
end
rescue ThreadError => e
puts "#{e.class}: #{e.message}:\n" + e.backtrace.join("\n") + "\n\n"
rescue Timeout::Error => e
#puts e.class
nil
end
end
end
end
threads.each{ |t| t.join }
Commenting synchronizing Block will cause the error to disappear but then then threads are not able to synchronize. I found some stuff on net, saying bug with ruby 192 , need changes in file prelude.rb and thread.c regarding MUTEX synchronization. But Under windows installation Unable to find file prelude.rb
Upvotes: 1
Views: 1253
Reputation: 84132
If a mutex is locked by a thread then an error will be raised if you try and lock it again from the same thread
This is exactly what you are doing, since synchronize
is just a convenience for method for locking the mutex, yielding to the block and then releasing the lock. I'm not sure what you're trying to do, but it feels to me like you might be trying to use mutexes for something other than their intended purposes.
Using threads and locks well is difficult to get right - you might want to look at celluloid for a different approach to concurrency.
Upvotes: 1