在使用None将默认参数设置为Python中的函数时，为什么会出现“ UnboundLocalError”？

Question

In the following function ,clc, whenever I fail to pass the third argument 'z', I want the fuction to automatically perform addition. (ie) If I pass only two arguments (150,200), I want to get the same result as I do when I pass (150,200,'a'). Unfortunately I am getting a traceback that 'UnboundLocalError: local variable 'r' referenced before assignment'. Can anyone please explain why I get this error? Thanks.

def clc(x,y,z=None):
    if (z=='a') or (z is None):
        r=x+y
    elif z=='s':
        r=x-y
    elif z=='m':
        r=x*y
    elif z=='d':
        r=x/y
    elif z=='mod':
        r=x%y
    elif z=='ex':
        r=x**y
    return r
while True:
    a=int(input('Enter a number:\t '))
    b=int(input('Enter a number:\t '))
    op=input('enter the operation:\t')
    answer=clc(a,b,op)
    print('the answer is',answer)
    query=input('If you want to continue, please press Enter\t')
    if len(query)>0:
        break

Edit: Thanks to PM2Ring for his explanation which cleared my error! This is his reply : 'If you just hit Enter at the 'enter the operation' prompt then op gets assigned the empty string '', it's not None. And then when you call clc(a,b,op) none of your if tests are True, and so r never gets assigned a value. The if statement I posted earlier handles that empty string case.'

Answer 1

Something tells me your input to op does not fit the bill for any of those if statements to get triggered, resulting in r never being defined or returned. One option would be to declare r = None outside the conditional block, so you get rid of that error.

However, if you're working with simple arithmetic operations (involving two operands only), I'd recommend keeping a dictionary of operations, and then invoking the correct one as needed.

import operator

_f = {'a' : operator.add, 's' : operator.sub, 'm' : operator.mul, 'd' : operator.div} 
def clc(x, y, op=None):
    return _f.get(op, operator.add)(x, y)

As an aside, as PM 2Ring remarked , if you want the default argument to come into play, don't pass anything to op .

---

Unfortunately, if you're looking to penalising invalid inputs, the function above needs modification, because it is highly tolerant to all invalid/missing inputs, defaulting to addition in each case. You can make a small change to get this to work, however, using the EAFP approach with exception handling.

_f.update({None : operator.add})
def clc(x, y, op=None):
    try:
       return _f[op](x, y)
    except KeyError:
        return "Invalid Operation"

Answer 2

The issue is the line op=input('enter the operation:\\t') . If you enter nothing, it still returns '' , an empty string, so for the z argument, you pass in '' . The interpreter runs through all of your if and elif statements, and none of them are satisfied, so it just returns r . However, r has not been assigned any value, so an exception is raised.

Instead, you should write something like this

def clc(x,y,z):
    if z=='a' or z=='':
        r=x+y
    elif z=='s':
        r=x-y
    elif z=='m':
        r=x*y
    elif z=='d':
        r=x/y
    elif z=='mod':
        r=x%y
    elif z=='ex':
        r=x**y
    else:
        return 'N/A'
    return r
while True:
    a=int(input('Enter a number:\t '))
    b=int(input('Enter a number:\t '))
    op=input('enter the operation:\t')
    answer=clc(a,b,op)
    print('the answer is',answer)
    query=input('If you want to continue, please press Enter\t')
    if len(query)>0:
        break