UnboundLocalError: local variable 'a' referenced before assignment
Question:
if execute the following code will show error message:
UnboundLocalError: local variable ‘a’ referenced before assignment
a = 220.0
b = 4300.0
c = 230.0/4300.0
def fun():
while (c > a/b):
a = a + 1
print a/b
if __name__ == '__main__':
fun()
but modify to :
a = 220.0
b = 4300.0
c = 230.0/4300.0
def fun():
aa = a
bb = b
while (c > aa/bb):
aa = aa + 1
print aa/bb
if __name__ == '__main__':
fun()
it will fine.
Any advice or pointers would be awesome. Thanks a lot!
Answers:
You can’t modify a global variable without using the global statement:
def fun():
global a
while (c > a/b):
a = a + 1
print a/b
As soon as python sees an assignment statement like a = a + 1 it thinks that the variable a is local variable and when the function is called the expression c > a/b is going to raise error because a is not defined yet.
if execute the following code will show error message:
UnboundLocalError: local variable ‘a’ referenced before assignment
a = 220.0
b = 4300.0
c = 230.0/4300.0
def fun():
while (c > a/b):
a = a + 1
print a/b
if __name__ == '__main__':
fun()
but modify to :
a = 220.0
b = 4300.0
c = 230.0/4300.0
def fun():
aa = a
bb = b
while (c > aa/bb):
aa = aa + 1
print aa/bb
if __name__ == '__main__':
fun()
it will fine.
Any advice or pointers would be awesome. Thanks a lot!
You can’t modify a global variable without using the global statement:
def fun():
global a
while (c > a/b):
a = a + 1
print a/b
As soon as python sees an assignment statement like a = a + 1 it thinks that the variable a is local variable and when the function is called the expression c > a/b is going to raise error because a is not defined yet.