Python 装饰器 - 斐波那契递归函数 - 意外输出

Question

I tried to play with using a decorator on a recursive function such as the one that calculates the nth Fibonacci number.

Below is the code I wrote:

class FibList:
# decorator will return a list of fibonacci numbers from 0 to n.
    def __init__(self, func):
        self.func = func

    def __call__(self, *args):
        return [self.func(x) for x in range(args[0])]

@FibList
def fibonacci(n):
    # function will return nth fibonacci number
    if n < 2:
        return n
    else:
        return fibonacci(n-2) + fibonacci(n-1)

if __name__ == "__main__":
    a = fibonacci(4)
    print(a)

The output for a looks like this: [0, 1, [0], [0, 0, 1]] . But, I am expecting the output looks like this: [0, 1, 1, 2]

I am really having a hard time understanding what is going on inside the decorated fibonacci function.

If anyone can help me to clarify such strange behavior, I will be much appreciated it. Thank you.

Answer 1

Your issue is caused by multiple recursion.

If you run the debugger and check each call of the recursive function, you will notice that the if-block returns the number correctly. That's why the first two numbers in your result list are correct. But as soon as it reaches the else-block, it will call the __call__ method of the Class with n-2 and n-1 respectively and thus create a new sublist inside of your result list.

When `n==2` the result would become:
fibonacci(n-2) -> __call__(self, 0) --> []
fibonacci(n-1) -> __call__(self, 1) --> [0]
return fibonacci(n-2) + fibonacci(n-1) --> [] + [0] = [0]

For n==3 you would get the last sublist [0,0,1]

You need to find a way to prevent this multi-recursive call.