CODEWITHSUNDEEP
python
Ankur Agarwal
Ankur Agarwal

Reputation: 24748

Error in the python script

I have this simple python script where myvar1 is accessible in generate() function but not in copy() function. Need help figuring out the error:

#!/usr/bin/python

import os, time

def Test(tcid,descr,iterations,filsz):

        def setup():
                print "entering set up\n"
                global myvar1, myvar2

                myvar1 = 1.0
                myvar2 = os.getcwd()

        def generate():
                print "entering generate\n"
                print "in generate", myvar1, myvar2

        def copy():

           print "in copy", myvar1, myvar2
           myvar1 += 5.0



        setup()
        generate()

        for loopcount in range(5):
           loopcount = loopcount + 1
           copy()



if __name__ == "__main__":
        Test('test','simple test',2,10)

Error:

Traceback (most recent call last): File "./pyerror.py", line 35, in Test('test','simple test',2,10) File "./pyerror.py", line 30, in Test copy() File "./pyerror.py", line 20, in copy print "in copy", myvar1, myvar2 UnboundLocalError: local variable 'myvar1' referenced before assignment

Upvotes: 1

Views: 176

Answers (2)

Jochen Ritzel
Jochen Ritzel

Reputation: 107588

You need global wherever you overwrite a global in a function. setup set myvar1 as a global but you used it as a unset local variable. Hence the "local variable 'myvar1' referenced before assignment"

def copy():
    global myvar1
    print "in copy", myvar1, myvar2
    myvar1 += 5.0

If you're reading a tutorial that suggests using global regularly then throw it away, burn it and start with a different one.

Upvotes: 2

Sven Marnach
Sven Marnach

Reputation: 601401

In the copy() function, myvar1 is not declared as global, but assigned to in the statement myvar1 += 5.0. This implicitly makes the myvar1 a local variable. The print statement in the first line of the function also tries to access a local variable with the name myvar1, but such a local variable does not exist yet.

Local variables are determined statically, i.e. at compile time. When the copy() is compiled by Python, myvar1 is marked as local variable for the whole function body. It may be instructive to look at the easiest code to trigger this error:

def f():
    print x
    x = 5
x = 3
f()

Upvotes: 2

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