[FIXED] why can't I declare a global variable inside an "exec" string

Issue

I am trying to declare and change a global variable in an exec string, like this:

ostr = "didn't work"
nstr = "worked"
def function():
    exec("global ostr; ostr = nstr")
    #global ostr; ostr = nstr
    print(ostr)
    lv='ostr' in globals()
    print(lv)
    ostr='asd'

function()

However, this errors out on the print statement with:

UnboundLocalError: local variable 'ostr' referenced before assignment

But, if I comment out the "exec" line and uncomment line after the exec statement, the code works fine.

How can I fix this error using "exec"? I want to declare variables global and modify those global variables inside an exec string and have those modifications visible in "function" on subsequent lines.

Solution

You have to declare that you are using global ostr in the function to be able to print it. This piece of code outputs

def function():
   global ostr
   exec("global ostr; ostr = nstr")
   #global ostr; ostr = nstr
   print(ostr)
   lv='ostr' in globals()
   print(lv)
   ostr='asd'

worked

True

Edit: Just realised the exec actually works with global variables, if you re-run your code and print(ostr) in the global main you will see it was changed.

ostr = "didn't work"
nstr = "worked"
def function():
    #global ostr
    exec("global ostr; ostr = nstr")

function()
print(ostr)

worked

Edit#2: Either declare ostr as a global variable before modifying it, or assign it to another local variable.

ostr = "didn't work"
nstr = "worked"
def function():
    exec("global ostr; ostr = nstr")
    #global ostr; ostr = nstr
    print(ostr)
    lv='ostr' in globals()
    print(lv)
        
function()

worked

True

Answered By - Anthony1223

Answer Checked By - Robin (PHPFixing Admin)