def test(a):
if a>1:
x=0
elif a<1:
y=1
else:
x=2
print(x)
return 0
Why test(2) is ok but test(0) will raise the following error?
local variable 'x' referenced before assignment
I guess when test(2) x was defined, but run test(0) x was not defined, but also want to know more about the cause
You pretty much answered it yourself. If a is 0, then the elif a<1 is true, so only y gets defined.
Because with test(0) the function test() only sets variable y but not variable x, and yet you are unconditionally printing x in the end, causing an UnboundLocalError exception.
Since you are not using y at all, it appears that it is actually a typo, and that you mean to set x to 1 there, so change your code to:
def test(a):
if a>1:
x=0
elif a<1:
x=1
else:
x=2
print(x)
return 0
That's because you're not initiating your variables in the function.
When the you have test(0), the second condition elif a<1: y=1 is satisfied. but then you try to print x which is not even defined and the interpreter doesn't know what it is.
def test(a):
...: if a>1:
...: x=0
...: elif a<1:
...: x=1
...: else:
...: x=2
...: print(x)
...: return 0
This will not raise the error.
