递归不终止于Python

Question

我试图写一个程序，找出给定的数字是否在斐波那契序列中，我不断得到不终止的递归，我不知道为什么。第17行似乎是个大问题。当我输入0或1时，我会得到想要的答案。我只是在寻找答案，我正在努力学习，所以仅仅告诉我答案对我没有多大帮助。

number = int(input("Enter your number:"))

def fib(n):
        if n == 0 or n == 1:
            return 1
        else:
            return (fib(n-1) + fib(n-2))
def isfib(number):
        n = 0
        if number < 0:
            print("Invalid input")
        elif number == fib(n):
            print("Number is in the sequence")
        elif number < fib(n):
            print("Number is not in the sequence")
        elif number > fib(n):
            n = n +1
            isfib(number) #where the problem occurs
isfib(number)

Answer 1

isfib（）函数的最后一个elif不是返回对自身的调用，也就是说，它正在调用自身，并且没有对该结果做任何操作。

Answer 2

你的fib功能错了（其中一项应该是n-2）

Answer 3

有很多小错误，所以我纠正了这些错误（我已经添加了一个更好的Fibonacci代码的实现与简单的添加太）：

number = int(input("Enter your number:"))

def fib(n): 
 if n == 0 or n == 1: return 1
 else:
  temp1=1
  temp=2
  temp3=0
  for z in range(n-2):
   temp3=temp
   temp+=temp1
   temp1=temp3
  return temp

def isfib(number): #it is ok not to return anything unless you need to stop the function in between
 done=0
 n=0
 while done!=1:
  if number < 0:
   print("Invalid input")
   done=1
  elif number == fib(n):
   print("Number is in the sequence")
   done=1
  elif number < fib(n):
   print("Number is not in the sequence")
   done=1
  elif number > fib(n):
   n = n +1
#i have used done instead of return to show the function can exit even if you dont return a value
#you can just 'return' instead of changing done variable and making the loop infinite
isfib(number)

既然你已经使用了很多递归，我猜你想只使用递归来做这件事。所以，代码是这样的：数字=int（输入（“输入你的数字：”））

def fib(n):
 if n == 0 or n == 1: return 1
 else: return (fib(n-1) + fib(n-2))
def isfib(number,n=0):
 if number < 0: print("Invalid input")
 elif number == fib(n): print("Number is in the sequence")
 elif number < fib(n): print("Number is not in the sequence")
 elif number > fib(n):
  n = n +1
  isfib(number,n)
isfib(number)

当然，经过测试，它是有效的（但我还是不推荐这样做：D）