Python 3 UnboundLocalError local variable referenced before assignment

Question

The following code gives the error UnboundLocalError  local variable  Var1  referenced before assignment   Var1   1 Var2   0 def function         if Var2    0 and Var1  gt  0          print  Result One       elif Var2    1 and Var1  gt  0          print  Result Two       elif Var1  lt  1          print  Result Three       Var1  - 1 function     How can I fix this  Thanks for any help

User · Answer

Why not simply return your calculated value and let the caller modify the global variable. It's not a good idea to manipulate a global variable within a function, as below:

Var1 = 1
Var2 = 0

def function(): 
    if Var2 == 0 and Var1 > 0:
        print("Result One")
    elif Var2 == 1 and Var1 > 0:
        print("Result Two")
    elif Var1 < 1:
        print("Result Three")
    return Var1 - 1

Var1 = function()

or even make local copies of the global variables and work with them and return the results which the caller can then assign appropriately

def function():
v1, v2 = Var1, Var2
# calculate using the local variables v1 & v2
return v1 - 1

Var1 = function()

User · Answer

I don t like this behavior  but this is how Python works  The question has already been answered by others  but for completeness  let me point out that Python 2 has more such quirks   def f x       return x  def main        print f 3      if  True           print  f for f in  1  2  3    main     Python 2 7 6 returns an error   Traceback  most recent call last     File  weird py   line 9  in  lt module gt      main     File  weird py   line 5  in main     print f 3  UnboundLocalError  local variable  f  referenced before assignment   Python sees the f is used as a local variable in  f for f in  1  2  3    and decides that it is also a local variable in f 3   You could add a global f statement   def f x       return x  def main        global f     print f 3      if  True           print  f for f in  1  2  3    main     It does work  however  f becomes 3 at the end    That is  print  f for f in  1  2  3   now changes the global variable f to 3  so it is not a function any more   Fortunately  it works fine in Python3 after adding the parentheses to print

User · Answer

This is because  even though Var1 exists  you re also using an assignment statement on the name Var1 inside of the function  Var1 -  1 at the bottom line   Naturally  this creates a variable inside the function s scope called Var1  truthfully  a -  or    will only update  reassign  an existing variable  but for reasons unknown  likely consistency in this context   Python treats it as an assignment   The Python interpreter sees this at module load time and decides  correctly so  that the global scope s Var1 should not be used inside the local scope  which leads to a problem when you try to reference the variable before it is locally assigned   Using global variables  outside of necessity  is usually frowned upon by Python developers  because it leads to confusing and problematic code  However  if you d like to use them to accomplish what your code is implying  you can simply add   global Var1  Var2   inside the top of your function  This will tell Python that you don t intend to define a Var1 or Var2 variable inside the function s local scope  The Python interpreter sees this at module load time and decides  correctly so  to look up any references to the aforementioned variables in the global scope   Some Resources   the Python website has a great explanation for this common issue  Python 3 offers a related nonlocal statement - check that out as well

User · Answer

You can fix this by passing parameters rather than relying on Globals  def function Var1  Var2        if Var2    0 and Var1  gt  0          print  Result One       elif Var2    1 and Var1  gt  0          print  Result Two       elif Var1  lt  1          print  Result Three       return Var1 - 1 function 1  1

User · Answer

If you set the value of a variable inside the function  python understands it as creating a local variable with that name  This local variable masks the global variable   In your case  Var1 is considered as a local variable  and it s used before being set  thus the error   To solve this problem  you can explicitly say it s a global by putting global Var1 in you function   Var1   1 Var2   0 def function        global Var1     if Var2    0 and Var1  gt  0          print  Result One       elif Var2    1 and Var1  gt  0          print  Result Two       elif Var1  lt  1          print  Result Three       Var1  - 1 function

[python] Python 3: UnboundLocalError: local variable referenced before assignment

Examples related to python

Examples related to function

Examples related to python-3.x