unbounded variable

[akbarza]

hi
in code:

'''
ref:https://www.bhavaniravi.com/python/advanced-python/unbound-variables-in-python
'''

def take_sum(a, b, c):
    return a+b+c

def main1():
    print(take_sum)    # output is as :<function take_sum at 0x1031c84c0>
    if "take_sum" not in  globals():
        print("take_sum is not in globals()")
    else:
        print("take_sum is in globals()")
    print(globals())
    print('-'*30)    


def main():
    if "take_sum" not in globals():
        take_sum = lambda x,y,z: x+y+z
    print(take_sum)
    

main1()
main()

output has the below error:

Error:
Traceback (most recent call last):
  File "D:\akb_python\akb_py_projects\unbounded_variable.py", line 25, in <module>
    main()
  File "D:\akb_python\akb_py_projects\unbounded_variable.py", line 21, in main
    print(take_sum)
UnboundLocalError: local variable 'take_sum' referenced before assignment

'''
I read the explanation given in the address in the docstring, but I did not understand. (I selected the thread subject from there).
what is the problem?plz, explain
thanks

[deanhystad]

Variables are "created" when Python code is parsed, not when it is executed. Before main() was ever called, the code for main "defined" a local vairable named "take_sum". You can see that if you look at the generated bytecode. Below I simplify the function to focus on the variable.

import dis


def main():
    if "take_sum" not in globals():
        take_sum = 1
    print(take_sum)


print(dis.dis(main))

Output:
             16 PRECALL                  0
             20 CALL                     0
             30 CONTAINS_OP              1
             32 POP_JUMP_FORWARD_IF_FALSE     2 (to 38)

  6          34 LOAD_CONST               2 (1)
             36 STORE_FAST               0 (take_sum)

  7     >>   38 LOAD_GLOBAL              3 (NULL + print)
             50 LOAD_FAST                0 (take_sum)
             52 PRECALL                  1
             56 CALL                     1
             66 POP_TOP
             68 LOAD_CONST               0 (None)
             70 RETURN_VALUE

Even if the comparison in line 5 (of the python code) causes execution to skip the assignment in line 6, main() still has a variable named (take_sum). It is part of the function's code.

main() creates (take_sum) because line 6 does an assignment to the variable. (take_sum) is a local variable because main() does not declare it as global.
Notice how the code changes if we declare take_sum as global.

import dis


def main():
    global take_sum
    if "take_sum" not in globals():
        take_sum = 1
    print(take_sum)


print(dis.dis(main))

Output:
  4           0 RESUME                   0

  6           2 LOAD_CONST               1 ('take_sum')
              4 LOAD_GLOBAL              1 (NULL + globals)
             16 PRECALL                  0
             20 CALL                     0
             30 CONTAINS_OP              1
             32 POP_JUMP_FORWARD_IF_FALSE     2 (to 38)

  7          34 LOAD_CONST               2 (1)
             36 STORE_GLOBAL             1 (take_sum)

  8     >>   38 LOAD_GLOBAL              5 (NULL + print)
             50 LOAD_GLOBAL              2 (take_sum)
             62 PRECALL                  1
             66 CALL                     1
             76 POP_TOP
             78 LOAD_CONST               0 (None)
             80 RETURN_VALUE

Now main() uses STORE_GLOBAL and LOAD_GLOBAL when referencing (take_sum).

To fix your code, you could modify main() to always assign something to take_sum:

def take_sum(a, b, c):
    return a+b+c
 
def main():
    if "take_sum" in globals():
        take_sum = globals()["take_sum"]
    else:
        take_sum = lambda x,y,z: x+y+z
    print(take_sum)

main()

Or you could declare take_sum as global

def take_sum(a, b, c):
    return a+b+c


def main():
    global take_sum
    if "take_sum" not in globals():
        take_sum = lambda x,y,z: x+y+z
    print(take_sum)
     

main()

[akbarza]

hi, thanks
I read your reply, although I am unfamiliar with dis module.so I did not understand the blue sections of your reply.
can I ask you why the error is not created in main1 function (in my Python code)?
thanks again

[deanhystad]

I am unfamiliar with dis module.so I did not understand the blue sections of your reply.

You can read all about the python disassembler here:
https://docs.python.org/3/library/dis.html

why the error is not created in main1 function

There are no assignments in main1, so main1 doesn't make a local variable. This would generate an error

def main1():
    print(take_sum) 

If you didn't have a function named "take_sum" in the global namespace. When accessing variables python first looks local, then enclosed, global and finally built-in.