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Disabling parallelism of subscribers in Python

asked 2020-07-03 04:31:03 -0600

omercan_s gravatar image

Hi all, I have observed that rospy is enabling the parallelism of subscriber callbacks which means that before one callback that receives a message from a topic finishes, another callback that receives a message from another topic can start execution. This might be ideal for most applications, but in my case it is not the desired behavior. To make it simple and general I am sharing the below code that would serve as an example:

#!/usr/bin/env python

import rospy
from std_msgs.msg import String

def callback1(msg):
    while(True):
        print("Callback1: %s" % (msg.data))
        rospy.sleep(1)

def callback2(msg):
    while(True):
        print("Callback2: %s" % (msg.data))
        rospy.sleep(1)

def main():
    rospy.init_node('foo')
    rospy.Subscriber('foo1', String, callback1, queue_size=1)
    rospy.Subscriber('foo2', String, callback2, queue_size=1)
    rospy.spin()

if __name__ == '__main__':
    main()

And the output is like:

Callback1: qwe
Callback2: asd
Callback1: qwe
Callback2: asd
Callback1: qwe

Is there any way to force a node to use a single thread for all of its callbacks which would mean a callback blocking the others from starting to execute until it finishes? Thanks in advance!

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1 Answer

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answered 2020-07-27 09:10:12 -0600

Bernat Gaston gravatar image

updated 2020-07-27 09:11:08 -0600

Callbacks are, by definition, parallel. I don't think there is a way to make them not parallel. What you can do is to set up a system of semaphores i.e. shared variables that block execution of the other thread until the one that is being processed finishes. Anyway, be careful when blocking callbacks, if you have a blocked callback and another message arrives, it will create another instance of your callback and you will have two blocked callbacks.

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Asked: 2020-07-03 04:31:03 -0600

Seen: 346 times

Last updated: Jul 27 '20