Hello I am newbie to Python. While learning I came across below code snippets in that I am unable to understand behaviour of the code.
Here is first case
#Case 1
x=1
def func():
x=5
print(x)
func()
5
Here is second case
#Case 2
x=1
def func():
print(x) #First print statement
x=5
print(x) #Second print statement
func()
UnboundLocalError: local variable 'x' referenced before assignment
The two cases are same except the second has one additional print statement. Why the first print statement of second snippet making the Python to throw an exception?
The x outside the function is there but x is also defined in both functions as a local variable and you must define it before using it. It doesn't mean that Python will use the x from outside the function first and then allow you to redefine x as a local variable and then use the local variable x in the rest of the function.
So the difference is really between:
def func():
x=5
print(x)
def func():
print(x) #First print statement
x=5
print(x) #Second print statement
In both functions x is a local variable but in the second function you're trying to use it before it's being defined in that function. Hence the error.
Related
Here is my code
def function():
global x
x = 5
print (x)
How can I access the local variable x inside the function without calling the function itself?
It's not possible. A variable inside a function will only be after the function is called, else it is non-existent. Look into how locallocal,non-local, and global variables work.
def function():
global x
x = 5
print(x)
function()
else you should just put x outside the function().
Just put x = 5 outside the function
So I get that
x = 5
def f():
print(x)
f()
print(x)
gives back 5 and 5.
I also get that
x = 5
def f():
x = 7
print(x)
f()
print(x)
gives back 7 and 5.
What is wrong with the following?
x = 5
def f():
if False:
x = 7
print(x)
else:
print(x)
f()
print(x)
I would guess that since the x=7 never happens I should get 5 and 5 again. Instead I get
UnboundLocalError: local variable 'x' referenced before assignment
Does python regard x as a local variable because in this indented block there is an assignment expression regardless if it is executed or not? What is the rule exactly?
When the function is defined python interprets x as a local variable since it's assigned inside the body of the function. During runtime, when you go into the else clause, the interpreter looks for a local variable x, which is unassigned.
If you want both x to refer to the same variable, you can add global x inside the body of the function, before its assignment to essentially tell python when I call x I'll be referring to the global-scope x.
If a name binding operation occurs anywhere within a code block, all uses of the name within the block are treated as references to the current block. This can lead to errors when a name is used within a block before it is bound. This rule is subtle. Python lacks declarations and allows name binding operations to occur anywhere within a code block. The local variables of a code block can be determined by scanning the entire text of the block for name binding operations.
You need to use global in your function f() like so:
x = 5
def f():
global x
if False:
x = 7
print(x)
else:
print(x)
f()
print(x)
This question already has answers here:
UnboundLocalError trying to use a variable (supposed to be global) that is (re)assigned (even after first use)
(14 answers)
Closed 5 years ago.
x=100
def fun2():
print x
x=10000
print x
fun2()
The above program showing local variable x reference before assignment. Why it is not printing
100
10000
x in the function is a local variable and can't access the other local variable you define first because they are in different scope.
Add global x to the start of your function or define x inside the function.
You appear to not know about variable scoping.
The variable x does not exist in the function scope.
You have to place global x before your print statement in order to access the global variable x.
x = 1 # Global x
def f():
x = 2 # function-local x
print(x) # prints 2
f()
print(x) # prints 1 because it uses the global x which remains unchanged
If you want that to work you need to specify inside the function that the x variable you are using is the one in the global scope by using the global keyword.
x=100
def fun2():
# Add this line
global x
print x
x=10000
print x
fun2()
Below code will print the value of x -> 100, as it is there in main scope #samba, but when you change the value of it doesn't work that way as it is not defined in the function.
x = 100
def fun2():
print(x)
fun2()
This doesn't work as the same way:
x = 100
def fun2():
print(x)
x = 1000
print(x)
fun2()
and through error:
UnboundLocalError: local variable 'x' referenced before assignment
x is a local variable and not initialised in function fun2().
You need to understand variable scoping here, Please check Global and Local variable scope
If you want to use it globally use global keyword in your function.
Because u assigned variable before function.
Just try this
def fun2():
x=100
print x
x=10000
print x
fun2()
It will output 100 and 1000
Okay, I am currently doing a project to make a blackjack game in python and I'm having some trouble. One of my issues is I dont know when to define a variable as global, specifically in functions with if statements. If I have a global variable outside the if statement, do I have to claim that the variable is global within the if statement as well? For example:
x = 5
def add():
global x <--- ?
x += 1
if x == 7:
global x <--- ?
x = 5
I'm pretty sure I need the "global x" at the 1st question mark, but what about at the second question mark? Would I still need to put a "global x" within my if statement if I wanted my if statement to update a global variable? Or does the global x at the beginning of the function make the x's inside the if statement global? Also, if I wanted to return x here, where should I do it?
Only one global statement is enough.
From docs:
The global statement is a declaration which holds for the entire
current code block.
x = 5
def add():
global x
x += 1
if x == 7:
x = 5
Also, if I wanted to return x here, where should I do it?
If you're using global in your function then the return x must come after the global x statement, if you didn't use any global statement and also didn't define any local variable x then you can return x anywhere in the function.
If you've defined a local variable x then return x must come after the definition.
Could someone explain why the following program fails:
def g(f):
for _ in range(10):
f()
def main():
x = 10
def f():
print x
x = x + 1
g(f)
if __name__ == '__main__':
main()
with the message:
Traceback (most recent call last):
File "a.py", line 13, in <module>
main()
File "a.py", line 10, in main
g(f)
File "a.py", line 3, in g
f()
File "a.py", line 8, in f
print x
UnboundLocalError: local variable 'x' referenced before assignment
But if I simply change the variable x to an array, it works:
def g(f):
for _ in range(10):
f()
def main():
x = [10]
def f():
print x[0]
x[0] = x[0] + 1
g(f)
if __name__ == '__main__':
main()
with the output
10
11
12
13
14
15
16
17
18
19
The reason I am confused is, if from f() it can't access x, why it becomes accessible if x is an array?
Thanks.
But this answer says the problem is with assigning to x. If that's it,
then printing it should work just fine, shouldn't it?
You have to understand the order in which things happen. Before your python code is even compiled and executed, something called a parser reads through the python code and checks the syntax. Another thing the parser does is mark variables as being local. When the parser sees an assignment in the code in a local scope, the variable on the lefthand side of the assignment is marked as local. At that point, nothing has even been compiled yet--let alone executed, and therefore no assignment takes place; the variable is merely marked as a local variable.
After the parser is finished, the code is compiled and executed. When execution reaches the print statement:
def main():
x = 10 #<---x in enclosing scope
def f():
print x #<-----
x = x + 1 #<-- x marked as local variable inside the function f()
the print statement looks like it should go ahead and print the x in the enclosing scope (the 'E' in the LEGB lookup process). However, because the parser previously marked x as a local variable inside f(), python does not proceed past the local scope (the 'L' in the LEGB lookup process) to lookup x. Because x has not been assigned to in the local scope at the time 'print x' executes, python spits out an error.
Note that even if the code where an assignment occurs will NEVER execute, the parser still marks the variable on the left of an assignment as a local variable. The parser has no idea about how things will execute, so it blindly searches for syntax errors and local variables throughout your file--even in code that can never execute. Here are some examples of that:
def dostuff ():
x = 10
def f():
print x
if False: #The body of the if will never execute...
a b c #...yet the parser finds a syntax error here
return f
f = dostuff()
f()
--output:--
File "1.py", line 8
a b c
^
SyntaxError: invalid syntax
The parser does the same thing when marking local variables:
def dostuff ():
x = 10
def f():
print x
if False: #The body of the if will never execute...
x = 0 #..yet the parser marks x as a local variable
return f
f = dostuff()
f()
Now look what happens when you execute that last program:
Traceback (most recent call last):
File "1.py", line 11, in <module>
f()
File "1.py", line 4, in f
print x
UnboundLocalError: local variable 'x' referenced before assignment
When the statement 'print x' executes, because the parser marked x as a local variable the lookup for x stops at the local scope.
That 'feature' is not unique to python--it happens in other languages too.
As for the array example, when you write:
x[0] = x[0] + 1
that tells python to go lookup up an array named x and assign something to its first element. Because there is no assignment to anything named x in the local scope, the parser does not mark x as a local variable.
The reason is in first example you used an assignment operation, x = x + 1, so when the functions was defined python thought that x is local variable. But when you actually called the function python failed to find any value for the x on the RHS locally, So raised an Error.
In your second example instead of assignment you simply changed a mutable object, so python will never raise any objection and will fetch x[0]'s value from the enclosing scope(actually it looks for it firstly in the enclosing scope, then global scope and finally in the builtins, but stops as soon as it was found).
In python 3x you can handle this using the nonlocal keyword and in py2x you can either pass the value to the inner function or use a function attribute.
Using function attribute:
def main():
main.x = 1
def f():
main.x = main.x + 1
print main.x
return f
main()() #prints 2
Passing the value explicitly:
def main():
x = 1
def f(x):
x = x + 1
print x
return x
x = f(x) #pass x and store the returned value back to x
main() #prints 2
Using nonlocal in py3x:
def main():
x = 1
def f():
nonlocal x
x = x + 1
print (x)
return f
main()() #prints 2
The problem is that the variable x is picked up by closure. When you try to assign to a variable that is picked up from the closure, python will complain unless you use the global or nonlocal1 keywords. In the case where you are using a list, you're not assigning to the name -- You can modify an object which got picked up in the closure, but you can't assign to it.
Basically, the error occurs at the print x line because when python parses the function, It sees that x is assigned to so it assumes x must be a local variable. When you get to the line print x, python tries to look up a local x but it isn't there. This can be seen by using dis.dis to inspect the bytecode. Here, python uses the LOAD_FAST instruction which is used for local variables rather than the LOAD_GLOBAL instruction which is used for non-local variables.
Normally, this would cause a NameError, but python tries to be a little more helpful by looking for x in func_closure or func_globals 2. If it finds x in one of those, it raises an UnboundLocalError instead to give you a better idea about what is happening -- You have a local variable which couldn't be found (isn't "bound").
1python3.x only
2python2.x -- On python3.x, those attributes have changed to __closure__ and __globals__ respectively
The problem is in the line
x = x + 1
This is the first time x being assigned in function f(), telling the compiler that x is a local name. It conflicts with the previous line print x, which can't find any previous assignment of the local x.
That's where your error UnboundLocalError: local variable 'x' referenced before assignment comes from.
Note that the error happens when compiler tries to figure out which object the x in print x refers to. So the print x doesn't executes.
Change it to
x[0] = x[0] + 1
No new name is added. So the compiler knows you are referring to the array outside f().