Why it shows error "Local variable 'output' might be referenced before assignment?

Question

def is_one_digit(v)
if type(v) == int:
if -10 < v < 10:
output = True
else:
output = False
return output

def check(v1, v2, v3):
if is_one_digit(v1) and is_one_digit(v2):
msg = "Good"
else:
msg = "Try again"
print(msg)

Answer 1

In the first method, output may be empty (unassigned) if type of v is int and v is not within the specified range. A better way to do this might be to just return the conditional statement, guaranteeing is_one_digit to return a valid boolean value under any circumstances.

def is_one_digit(v):
return ((type(v)==int) and (-10 < v < 10))

Answer 2

Neethu Nath Python supports short-circuiting. For 'and', the second argument is only evaluated if the first one is true.

https://docs.python.org/3/library/stdtypes.html#boolean-operations-and-or-not

https://code.sololearn.com/cLXQ08VhE2Dy/?ref=app

Answer 3

Issue: If v == type Integer, but not 1 digit, what value has output? I think undefined.
So try one new line:

def is_one_digit(v) :
output = False
.
.
.

Answer 4

Or, 2. way:

if type(v) == int:
if - 10 < v <10:
output = True
else:
output = False
else:
output = False
return output

Answer 5

I tried that. It shows error as the input can be a non number.

Answer 6

Hatsy Rei Let me try