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Python interpreter sequence with a function [duplicate]

Suppose I have a Python function foo which takes a default argument, where that default is set to some global variable. If I now change that global variable before calling the function, the default argument is still set to the original value of that global variable.
For example:
x = 1
def foo(a=x):
print a
x = 2
foo()
This prints 1, instead of 2.
How should I be writing my code, so that I can change this global variable, and have it update this default argument?

A default variable is only evaluated and set once. So Python makes a copy of the reference and from then on, it always passes that reference as default value. No re-evaluation is done.
You can however solve this by using another object as default object, and then use an if statement to substitute it accordingly. Something like:
the_default = object()
x = 1
def foo(a = the_default):
if a is the_default:
a = x
print a
x = 2
foo()
Note that we use is to perform reference equality. So we check if it is indeed the default_object. You should not use the the_default object somewhere else in your code.
In many Python code, they use None as a default (and thus reduce the number of objects the construct). For instance:
def foo(a = None):
if a is None:
a = x
print a
Note however that if you do that, your program cannot make a distinction between a user calling foo() and foo(None) whereas using something as the_object makes it harder for a user to obtain a reference to that object. This can be useful if None would be a valid candidate as well: if you want foo(None) to print 'None' and not x.

Why can a function change an objects variable but not a variable

Here's what I don't understand, if I made a function that would change the value of a variable, this would only save it in the function, so it won't change the global variable.
var = 10
def change_var(variable):
variable += 1
change_var(var)
print(var)
________________________
10
However, when I use the variable of a object (I'm not sure what this is called), it works completely fine. This just doesn't makes sense to me that one works but the other doesn't.
This is what I mean
class foo():
def __init__(self, var):
self.var = var
object_ = foo(10)
def change_var(object_):
object_.var += 1
change_var(object_)
print(object_.var)
________________________
11
I want an explanation on why one works but not the other

Python passes variables by Value, but objects by Reference.
So if you modify a variable, you modify your local copy, not the original; if you modify an object, you modify the original.

Saving and Loading a Class to a File [duplicate]

This question already has answers here:
Using global variables in a function
(25 answers)
Closed 5 months ago.
I know I should avoid using global variables in the first place due to confusion like this, but if I were to use them, is the following a valid way to go about using them? (I am trying to call the global copy of a variable created in a separate function.)
x = "somevalue"
def func_A ():
global x
# Do things to x
return x
def func_B():
x = func_A()
# Do things
return x
func_A()
func_B()
Does the x that the second function uses have the same value of the global copy of x that func_a uses and modifies? When calling the functions after definition, does order matter?

If you want to simply access a global variable you just use its name. However to change its value you need to use the global keyword.
E.g.
global someVar
someVar = 55
This would change the value of the global variable to 55. Otherwise it would just assign 55 to a local variable.
The order of function definition listings doesn't matter (assuming they don't refer to each other in some way), the order they are called does.

Within a Python scope, any assignment to a variable not already declared within that scope creates a new local variable unless that variable is declared earlier in the function as referring to a globally scoped variable with the keyword global.
Let's look at a modified version of your pseudocode to see what happens:
# Here, we're creating a variable 'x', in the __main__ scope.
x = 'None!'
def func_A():
# The below declaration lets the function know that we
# mean the global 'x' when we refer to that variable, not
# any local one
global x
x = 'A'
return x
def func_B():
# Here, we are somewhat mislead. We're actually involving two different
# variables named 'x'. One is local to func_B, the other is global.
# By calling func_A(), we do two things: we're reassigning the value
# of the GLOBAL x as part of func_A, and then taking that same value
# since it's returned by func_A, and assigning it to a LOCAL variable
# named 'x'.
x = func_A() # look at this as: x_local = func_A()
# Here, we're assigning the value of 'B' to the LOCAL x.
x = 'B' # look at this as: x_local = 'B'
return x # look at this as: return x_local
In fact, you could rewrite all of func_B with the variable named x_local and it would work identically.
The order matters only as far as the order in which your functions do operations that change the value of the global x. Thus in our example, order doesn't matter, since func_B calls func_A. In this example, order does matter:
def a():
global foo
foo = 'A'
def b():
global foo
foo = 'B'
b()
a()
print foo
# prints 'A' because a() was the last function to modify 'foo'.
Note that global is only required to modify global objects. You can still access them from within a function without declaring global.
Thus, we have:
x = 5
def access_only():
return x
# This returns whatever the global value of 'x' is
def modify():
global x
x = 'modified'
return x
# This function makes the global 'x' equal to 'modified', and then returns that value
def create_locally():
x = 'local!'
return x
# This function creates a new local variable named 'x', and sets it as 'local',
# and returns that. The global 'x' is untouched.
Note the difference between create_locally and access_only -- access_only is accessing the global x despite not calling global, and even though create_locally doesn't use global either, it creates a local copy since it's assigning a value.
The confusion here is why you shouldn't use global variables.

You can directly access a global variable inside a function. If you want to change the value of that global variable, use "global variable_name". See the following example:
var = 1
def global_var_change():
global var
var = "value changed"
global_var_change() #call the function for changes
print var
Generally speaking, this is not a good programming practice. By breaking namespace logic, code can become difficult to understand and debug.

As others have noted, you need to declare a variable global in a function when you want that function to be able to modify the global variable. If you only want to access it, then you don't need global.
To go into a bit more detail on that, what "modify" means is this: if you want to re-bind the global name so it points to a different object, the name must be declared global in the function.
Many operations that modify (mutate) an object do not re-bind the global name to point to a different object, and so they are all valid without declaring the name global in the function.
d = {}
l = []
o = type("object", (object,), {})()
def valid(): # these are all valid without declaring any names global!
d[0] = 1 # changes what's in d, but d still points to the same object
d[0] += 1 # ditto
d.clear() # ditto! d is now empty but it`s still the same object!
l.append(0) # l is still the same list but has an additional member
o.test = 1 # creating new attribute on o, but o is still the same object

Here is one case that caught me out, using a global as a default value of a parameter.
globVar = None # initialize value of global variable
def func(param = globVar): # use globVar as default value for param
print 'param =', param, 'globVar =', globVar # display values
def test():
global globVar
globVar = 42 # change value of global
func()
test()
=========
output: param = None, globVar = 42
I had expected param to have a value of 42. Surprise. Python 2.7 evaluated the value of globVar when it first parsed the function func. Changing the value of globVar did not affect the default value assigned to param. Delaying the evaluation, as in the following, worked as I needed it to.
def func(param = eval('globVar')): # this seems to work
print 'param =', param, 'globVar =', globVar # display values
Or, if you want to be safe,
def func(param = None)):
if param == None:
param = globVar
print 'param =', param, 'globVar =', globVar # display values

You must use the global declaration when you wish to alter the value assigned to a global variable.
You do not need it to read from a global variable. Note that calling a method on an object (even if it alters the data within that object) does not alter the value of the variable holding that object (absent reflective magic).

Acsess from function to non-global variables [duplicate]

This question already has answers here:
Why isn't the 'global' keyword needed to access a global variable?
(11 answers)
Closed 6 months ago.
This is a weird behavior.
Try this :
rep_i=0
print "rep_i is" , rep_i
def test():
global rep_i #without Global this gives error but list , dict , and others don't
if rep_i==0:
print "Testing Integer %s" % rep_i
rep_i=1
return "Done"
rep_lst=[1,2,3]
def test2():
if rep_lst[0]==1:
print "Testing List %s" % rep_lst
return "Done"
if __name__=="__main__":
test()
test2()
Why list do not need to declare global? are they automatically global?
I find it really weird, I use list most of the time and I don't even use global at all to us them as global...

It isn't automatically global.
However, there's a difference between rep_i=1 and rep_lst[0]=1 - the former rebinds the name rep_i, so global is needed to prevent creation of a local slot of the same name. In the latter case, you're just modifying an existing, global object, which is found by regular name lookup (changing a list entry is like calling a member function on the list, it's not a name rebinding).
To test it out, try assigning rep_lst=[] in test2 (i.e. set it to a fresh list). Unless you declare rep_lst global, the effects won't be visible outside test2 because a local slot of the same name is created and shadows the global slot.

You only need to use global if you are assigning to the global name. Without global, an assignment creates a new local.
There's nothing special about how global applies to a list—global simply influences scope and name resolution.

There is an error in python called UnboundLocalError which often confuses newcomers. The confusing thing is: future assignment does change the way a variable is looked up.
When the interpreter sees a variable name for the first time, it looks ahead to the end of current code block, and if you don't have an assignment to it anywhere within the same block of code, the interpreter considers it global. If you do, however, then it is considered local, and any reference to it before assignment generates an UnboundLocalError. That's the error you got. That's why you need to declare global rep_i. If you did not assign rep_i, you wouldn't need this line.
Also, this has nothing to do with variable type. Also, assigning or appending an item to the list (which you probably meant to do, but did not) is not assignment of the list itself, it is essentially calling a method on a list object, which is different from assignment: assignment creates a new object (possibly under a name that already exists), while manipulating a list just changes an existing list.
You can try:
In [1]: # It won't work with small integers, as they are cached singletons in CPython
In [2]: a = 123123
In [3]: id (a)
Out[3]: 9116848
In [4]: a = 123123
In [5]: id(a)
Out[5]: 9116740
In [6]: # See, it changed
In [7]: # Now with lists
In [8]: l = [1,2,3]
In [9]: id(l)
Out[9]: 19885792
In [10]: l[1] = 2
In [11]: id(l)
Out[11]: 19885792
In [12]: # See, it is the same
In [13]: # But if i reassign the list, even to the same value
In [14]: l = [2,2,3]
In [15]: id(l)
Out[15]: 19884272

Here's an example that demonstrates that a non list/dict variable is available in a subroutine, and the problem is, as everyone says, the act of rebinding in your original code sample:
x = 1
def test():
y = x + 1
print y
test()
You'll see this prints out 2, despite x not being declared global.

If you had assigned a new value to rep_lst inside of test2 (not just to one of its elements, as you did) it would not work without the global flag. In Python, if you do not assign to a variable inside a function it will look for that variable in in more global scopes until it finds it.
For example, in this code segment I define the list both globally and inside of example(). Since the variable in example() is closer in scope to example2() than the global one is, it is what will be used.
x = ["out"]
def example():
x = ["in"]
def example2():
print x # will print ["in"]
This has nothing to do with lists, but is the behaviour of any variable in Python.

Correct Use Of Global Variables In Python 3

Which is the correct use of global variables in Python 3?:
1) Stating global VAR_NAME once in the core script (not within a function) and then simply referring to the variable as VAR_NAME everywhere else
2) Stating global VAR_NAME once within every function that uses the global variable and then simply referring to the variable as VAR_NAME for the rest of the function and within the core script itself

In the first case the global keyword is pointless, so that is not correct. Defining a variable on the module level makes it a global variable, you don't need the global keyword.
The second example is correct usage.
However, the most common usage for global variables are without using the global keyword anywhere. The global keyword is needed only if you want to reassign the global variables in the function/method.

You need to use the global keyword in a function if you use the global variable in a way that would otherwise be interpreted as an assignment to a local variable. Without the global keyword, you will create a local variable that hides the global in the scope of the function.
Here are a few examples:
global_var = 1
def example1():
# global keyword is not needed, local_var will be set to 1.
local_var = global_var
def example2():
# global keyword is needed, if you want to set global_var,
# otherwise you will create a local variable.
global_var = 2
def example3():
# Without using the global keyword, this is an error.
# It's an attempt to reference a local variable that has not been declared.
global_var += 1

"in a way that would otherwise be interpreted as an assignment to a local variable" --- yes, but here is a subtle detail:
------------------- error: local variable 'c' referenced before assignment
def work():
c += 3
c = 0
work()
print(c)
------------------- error: local variable 'c' referenced before assignment
c = 0
def work():
c += 3
work()
print(c)
------------------- prints [3]
def work():
c.append(3)
c = []
work()
print(c)
------------------- prints [3]
c = []
def work():
c.append(3)
work()
print(c)

The main difference between the first two cases and the next two cases in the above answer would have to be the fact that the list is mutable. For cases like a = 1 a pointer points to the location where 1 is and when you say a = 2 the pointer shifts.
For the case of mutable objects a memory location is allotted and when methods like append are used changes occur to the memory location itself and so the value the mutable references is changed globally.
Now the big question is as to how the function knows the variable we are modifying is a global one or local one because it seems we can modify the global variable if its mutable and we cannot if its non mutable (The function also does not recognize this as the global variable)