CODEWITHSUNDEEP
pythonpython-2.7
Ruthvik Vaila
Ruthvik Vaila

Reputation: 537

accessing a variable from outside the function in python

#!/usr/bin/env python
import roslib
import rospy
import time
from nav_msgs.msg import Odometry 

def position_callback(data):
    global q2
    q2=data.pose.pose.position.x
    q1=data.pose.pose.position.y
    q3=data.pose.pose.position.z


def position():      
    rospy.init_node('position', anonymous=True)  #initialize the node"
    rospy.Subscriber("odom", Odometry, position_callback)

if __name__ == '__main__':

    try:
        position()
        print q2
        rospy.spin()
    except rospy.ROSInterruptException: pass

the error i get is like this:

print q2
NameError: global name 'q2' is not defined

I defined q2 as global variable already.

Upvotes: 1

Views: 5264

Answers (2)

falsetru
falsetru

Reputation: 368894

Declaring q2 as a global variable does make the global variable exist.

Actually calling the function and execution of the assignment statement q2 = ... cause the creation of the variable. Until then, the code cannot access the variable.

position function does not call the position_callback, but pass it to rospy.Subscriber (which probably register the callback function, and not call it directly).

Initialize q2 if you want to access the variable before it is set.

q2 = None

def position_callback(data):
    global q2
    q2 = data.pose.pose.position.x
    q1 = data.pose.pose.position.y
    q3 = data.pose.pose.position.z

Upvotes: 2

Yair Daon
Yair Daon

Reputation: 1123

You never initialize q2. It cannot have a value. Try to define it in global scope - after the imports. Then call it iniside the functionposition().

Upvotes: 0

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