现在的问题是，你的返回是是函数的循环中。因此在第一次迭代之后，将已经停止，并返回所述第一值：1.除非 N 为0，在这种情况下，功能是由以返回 0 本身，如果 N 1，当循环不会重复，甚至一次也没有返回正在被执行（因此无返回值）。

要解决这个问题，只要移动返回是的循环之外。

替代实施

随着KebertX的例子，这里是一个解决方案，我会亲自做在Python。当然，如果你要处理许多斐波纳契值，你甚至可能要合并这两个解决方案，并为数字的高速缓存。

 高清F（N）：
    的a，b = 0，1
    对于i在范围（0，n）的：
        的a，b = B，A + B
    返回
 

I am interested in a iterative algorithm for Fibonacci numbers, so I found the formula on wiki...it looks straight forward so I tried it in Python...it doesn't have a problem compiling and formula looks right...not sure why its giving the wrong output...did I not implement it right ?

def fib (n): 
    if( n == 0):
        return 0
    else:
        x = 0
        y = 1
        for i in range(1,n):
            z = (x + y)
            x = y
            y = z
            return y

for i in range(10):
    print (fib(i))

output

0
None
1
1
1
1
1
1
1
1
1
1

The problem is that your return y is within the loop of your function. So after the first iteration, it will already stop and return the first value: 1. Except when n is 0, in which case the function is made to return 0 itself, and in case n is 1, when the for loop will not iterate even once, and no return is being execute (hence the None return value).

To fix this, just move the return y outside of the loop.

Alternative implementation

Following KebertX’s example, here is a solution I would personally make in Python. Of course, if you were to process many Fibonacci values, you might even want to combine those two solutions and create a cache for the numbers.

def f(n):
    a, b = 0, 1
    for i in range(0, n):
        a, b = b, a + b
    return a