Question

I’m learning decorators and here I’m trying to change the below to decorator pattern.

def invert(x):
    return 1/x

print invert(5)

can be changed using decorators.

def safe(fun, *args):
    if args[0]!=0:
        return fun(*args)
    else:
        "Division by 0"

def invert(x):
    return 1/x

print safe(invert, 5) 

Using @wapper syntax,

def safe(fun, *args):
    if args[0]!=0:
        return fun(*args)
    else:
        "Division by 0"
@safe
def invert(x):
    return 1/x

print invert(5)

The above code gives error IndexError: tuple index out of range. I’m trying to understand what makes it wrong and how to correct it.

Answer 1

Your main problem is that a decorator needs to return a function, not a value. The decorator replaces the defined function with a new one based on it.

Beyond that, you do not have a return statement in the else block, merely a string literal. You probably meant to return that.

def safe(fun):
    def f(*args):
        if not args[0] == 0:
            return fun(*args)
        else:
            return "Division by 0"
    return f

Just as a note, I presume this is for the purposes of the exercise only, but errors should not pass silently in Python – the behaviour of emitting an exception is far more useful in general than returning a string on error. In fact, to implement this more naturally for Python you want to follow a ask for forgiveness, not permission mantra:

def safe(fun):
    def f(*args):
        try:
            return fun(*args)
        except ZeroDivisionError:
            return "Division by 0"
    return f

Also, be careful of sweeping statement names like ‘safe’ – other exceptions could still be thrown.