Python Decorator - Fibonacci Recursive Function - Unexpected Output-CodePudding

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.

CodePudding user response：

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.