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I got the impression that a node per default has a namespace matching the node name.
Mini example, I run this node:
import rospy
rospy.init_node('tester')
rospy.set_param('foo', 1)
rospy.spin()
I then run rosparam list, I get:
/foo
/rosdistro
/roslaunch/uris/host_knxps13_local__35409
/rosversion /run_id
Why don't I get:
/tester/foo
Why is the namespace of the node / and not /tester?
Short answer. rospy nodes do not have a direct equivalent to roscpp ros::NodeHandle.
See (almost) duplicate question and answer - including way to resolve names relative to a rospy node's name - here.
Edit:
Yes, for rospy, in order to have rosparam list generate this:
/tester/foo
/rosdistro
/roslaunch/uris/host_knxps13_local__35409
/rosversion
/run_id
You will need to set the foo as a private parameter of the tester rospy node:
import rospy
rospy.init_node('tester')
rospy.set_param('~foo', 1)
rospy.spin()
Additional info:
http://wiki.ros.org/rospy/Overview/Pa...
http://wiki.ros.org/rospy_tutorials/T...
http://wiki.ros.org/Names#Resolving
@knxa That is correct. When you call rospy.resolve_name('~bar'), you are telling ROS, "Give me the global name of relative name ~bar" - in this case /tester/bar. If you were to call
rospy.set_param('foo', 1), then rospy.resolve_name('foo') would return /foo...
since / is the namespace in which the relative parameter name foo was declared. If you were to launch tester in another namespace ns1 (via roslaunch), then rospy.resolve_name('~bar') would return /ns1/tester/bar and rospy.resolve_name('food') would return /ns1/foo
Asked: 2018-02-16 08:49:29 -0600
Seen: 683 times
Last updated: Feb 16 '18
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