Re: Variable scope inside and outside functions

Jacob Kruger <jacob.kruger.work@gmail.com> writes:    if you execute

do_it function, it will print out the new value assigned to the
dt_expiry variable inside that function, but if you then again check the
value of the dt_expiry variable afterwards, it's reverted to the 1970...
value?

  I can only guess that after "do_it" you might have inadvertently
  restarted the interpreter.