CODEWITHSUNDEEP
python-3.x
Renxing Liu
Renxing Liu

Reputation: 101

Python: TypeError: 'int' object is not iterable

I'm tackling this following question:

Write a function is_fib(n) that returns True if n is a Fibonacci number, and False otherwise.

This is my code:

def is_fib(n):
    def fib(x):
        if x == 0:
            return 0
        elif x == 1:
            return 1
        else:
            return fib(x-1) + fib(x-2)
    for a in n:
        if fib(a) == n:
            result = True
            break
        else:
            result = False
    return result

Running this give rise to:

TypeError: 'int' object is not iterable.

I have been staring at the code for half an hour. Any help is greatly appreciated.

Upvotes: 1

Views: 1518

Answers (3)

Renxing Liu
Renxing Liu

Reputation: 101

Firstly thanks to the two guys that helped me. However, for Yoav's edition, python will run into an error when n is a really big number.

This is my new and improved version.

def is_fib(n):
    if n < 0:
        return False
    else:
        fib_0 = 0
        fib_1 = 1
        if n == fib_0 or n == fib_1:
            return True
        else:
            for a in range(2,n+2):
                fib = fib_0 + fib_1
                fib_0,fib_1 = fib_1,fib
                if fib >= n:
                    break
            return fib == n

Upvotes: 0

Yoav Glazner
Yoav Glazner

Reputation: 8041

As jozefg said you are missing range(n)

also notice that you need range(n+2) to cover all cases

def is_fib(n):
    def fib(x):
        if x == 0:
            return 0
        elif x == 1:
            return 1
        else:
            return fib(x-1) + fib(x-2)
    for a in range(n+2):
        if fib(a) == n:
            return True


    return False

print(is_fib(3))

Upvotes: 0

daniel gratzer
daniel gratzer

Reputation: 53911

I think you mean

for a in range(n)

not

for a in n

Upvotes: 1

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