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().
Related
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)
I'm reading the python reference name resolution, which reads
A class definition is an executable statement that may use and define names. These references follow the normal rules for name resolution with an exception that unbound local variables are looked up in the global namespace.
Based on that, I would expect the following code
x = 10
def f():
x = 5
class Test:
y = x
return Test
print(f().y)
to print 10, however it prints 5. is this a mistake in the reference, or am I misunderstanding something?
In this case, 'normal' rules apply:
x = 'global'
def f():
x = 'defined in f'
class Test:
print(x) # not local, normal rules apply
f()
# defined in f
In this second case, we would expect an UnboundLocalError: local variable 'x' referenced before assignment if we were inside a function:
x = 'global'
def f():
x = 'defined in f'
class Test:
print(x) # unbound local at this time
x = 'assigned in Test'
print(x)
But for the first print(x), x will be taken from the global namespace:
f()
# global
# assigned in Test
I think #khelwood gave the answer. The value in the variable:
Test.y
is an integer that you define, but you never give the x = 10 to the function.
Maybe what you want is actually:
x = 10
def f(x=5):
class Test:
y = x
return Test
print(f().y) #print default 5
print(f(x).y)
The last line print 10 since x is given to the function, and therefore the class set y as x
You can see the exception by wrapping f in another function that defines a local variable x:
x = 10
def g():
x = 7
def f():
class Test:
y = x
return Test
return f()
print(g().y)
Now the output is 7, as the lookup of x in the class statement bypasses the local scope of g to find the global value of x.
In your code, because class establishes a new namespace, but not a new scope, the value of x is taken from the current local scope, that of f.
This question already has answers here:
When is the existence of nonlocal variables checked?
(2 answers)
Closed 4 years ago.
previously I thought that when we define a function, the function can be wrong, but python will not check it until it got executed:
x = 100
def f():
x = 1/0
return x
print(x)
# >>> 100
however, when I was learning the nonlocal statement
x = 100
def f():
def g():
nonlocal x
x = x * 99
return x
return g
print(x)
# >>> SyntaxError: no binding for nonlocal 'x' found
It got checked out even if the function is not executed.
Is there anywhere I can find the official explanation for this?
Additional for variable bounding situation:
x = 100
def f():
global x
global xx
x = 99
return x
print(f())
# >>> 99
print(x)
# >>> 99
it seemed totally fine, if I global some variable that does not exist at all?
And it doesn't even bring any error even if I execute this function?
this part is moved to a new individual question:
Why am I able to global a non-existing varlable in python
The nonlocal checks the nearest enclosing scope, excluding globals (that is, module level variables). That is, your f() function should declare a x for it to work, as nonlocal can't see the global x = 100 variable.
See https://docs.python.org/3/reference/simple_stmts.html#the-global-statement.
As to why the error is raised without running the function, it is because the variable names are bound at compile-time, so it doesn't matter that you don't use this function at all. See
https://docs.python.org/3/reference/executionmodel.html#resolution-of-names
However, global has different behaviour. Just like nonlocal if the global x already exists, it is used instead of the local one. However, if it doesn't, it means "if I create a variable named x, create it on the global scope, instead of the function scope". So, in your example
x = 100
def f():
global x
global xx
x = 99
xx = 123
return x
print(f()) # 99
print(x) # 99
print(xx) # 123
a xx variable has appeared in the global namespace. It is only a hint to the compiler though, so if you declared global xx without assigning it, and try to print(xx) later, you still get a NameError for using an undefined variable
from random import randint
shifts = [4, 4.2, 5, 6, 7]
days_names = ['Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday']
workers_names = ['Itai', 'Or', 'Reut', 'Kuka', 'Aviel']
counter = 1
def shift_arrange(worker):
for day in days.values():
counter+=1
global avilable_shifts
avilable_shifts = check_avilable_shifts(day)
if not random_shifte_selector(worker,day): soft_reset(worker)
I set counter as a global variable, and when i try to run this code i get the local variable error:
Traceback (most recent call last):
File "C:\Or\mypy\shift creator\shift cretor.py", line 144, in <module>
for w in workers.values(): shift_arrange(w)
File "C:\Or\mypy\shift creator\shift cretor.py", line 105, in shift_arrange
counter+=1
UnboundLocalError: local variable 'counter' referenced before assignmen
I saw some guy ask this question here, he deleted his pyc file or something(i don't know what is it) and its work fine. Why this is happen? Its not happen to other variables in the program.
Thanks, Or
You need to declare a global variable
def shift_arrange(worker):
global counter
for day in days.values():
counter+=1
...
Since you modify counter in that scope, python treats it as a local variable, unless you declare it as global. If you only need to read it, that isn't necessary.
Consider the following:
This works:
c = 0
def f():
print c
f()
While this does not:
c = 0
def f():
print c
c = 1
f()
While this does:
c = 0
def f():
global c
print c
c = 1
f()
print c # prints 1, f() modified the global value
There's a lot to look up here, but in general python has names, and things that are assigned to them. The name could be either in the local or global scope.
For reads, python looks through local scope, then through global scope, and uses the first one it finds (or error if it doesn't).
For writes... python needs to know where to put it. Normally, what it would do is look in the local scope, and if it's not there, create a variable there and assign the value. This would hide the global variable. You could have it also look in globals and use that one if it exists - but that could be undesirable & unexpected. So, you need a way to tell python to use a global variable instead if it exists (and then, it will create if it doesn't).
This leads to some odd behaviour sometimes as well. In addition to the earlier answer...
c = 0
# Passes. We are just looking up the global variable.
def f1(x):
return x + c
# Passes, but may not be as expected. Sets a local variable c to a value, does not
# modify the global one.
def f2(x):
c = x
# Correct way to do the above; now it sets the global variable.
def f3(x):
global c
c = x
# What if you mix them?
def f4(x):
c = c + x
# This fails. What happens is that first it sees that it's writing to c, and it's
# not marked global, so it assigns c to the local space. Then it tries to dereference
# it. Since we've marked it local, it masks the global one, and since it doesn't
# have a value, it throws an error. c += x works the same way and also fails.
# However, this works, though is just as equally wrong:
def f5(x):
d = c
c = d + x
# This one works, because we copy the value of the global c into the local d.
# Then, the second line assigns a local c to addition of local d and x.
# But does not update the global.
# Each line is separate though:
def f6(x):
d = c
c = c + 1
# You might think that d=c already made c local for the whole function. But it doesn't
# work like that. The first line just looks up the value for c, which it finds
# globally, but then forgets about it - it cares about the object c is the name of,
# not the name c itself. The second line still fails.
In python (tested on 2.7.6) all variables are
statically bound to a scope at compile time. This process is well
described in http://www.python.org/dev/peps/pep-0227/ and
http://docs.python.org/2.7/reference/executionmodel.html
It is explicitly stated that "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."
A function is a code block so the following code with fail because x
is assigned after its use (so at compile time it is defined local
because it is assigned somewhere in the function, but at execution
time, it is used before being bound).
x = 1
def f():
print x
x = 2
print x
>>> f()
Traceback (most recent call last):
File "<pyshell#46>", line 1, in <module>
f()
File "<pyshell#45>", line 2, in f
print x
UnboundLocalError: local variable 'x' referenced before assignment
A class is also a code block, so we should observe exactly the
same behavior. But this is not what I observe.
Look at this example:
x = 1
class C():
y = x + 10
x = 2
def __init__(self):
print C.y
>>> C.x
2
>>> C.y
11
>>> C()
11
<__main__.C instance at 0x00000000027CC9C8>
As the class definition is a code block, any assignment within this
block should make the variable local. So x should be local to the
class C, so y = x + 10 should result in an UnboundLocalError.
Why there is not such error?
Yes - it seems that the documentation is rather misleading. A class definition doesn't actually work quite the same as other normal blocks:
global_one = 0
class A(object):
x = global_one + 10
global_one = 100
y = global_one + 20
del global_one
z = global_one + 30
a = A()
print a.x, a.y, a.z, global_one
results in: 10, 120, 30, 0
if you try the same thing with a function, you get an UnboundLocalError on your first access of global_one.
The reason for this is that class definitions as normal have access to the parent scope, however, all name assignments do NOT modify a local scope, but in fact are captured into the class's data attributes dictionary. There are hints about this in the documentation, but it's certainly not obvious.