Python为什么我定义的函数（其输出是列表）不断收到nonetype错误

Question

I define a function which output is to print a list. then i want to compare that list with len(l2) which is some other list in my code but i keep receiving this error that my defined function's output is a NoneType and it cant be compared with integers how can i solve this problem? Im new to programming so please answer simply

l = []
for i in range(0, 3):
    x = int(input())
    l.append(x)
def prime_counter(n):
    b = 1
    l1 = []
    while b <= n:
        k = 0
        if n % b == 0:
            j = 1
            while j <= b:
                if b % j == 0:
                    k = k + 1
                j = j + 1
            if k == 2:
                l1.append(b)
        b = b + 1
for i in range(0, len(l)):
    l2 = []
    if len(prime_counter(l[i])) > len(l2):
        l2.extend(prime_counter(l[i]))
print(l2)

Answer 1

When you don't have any return statement in a function, None is returned automatically. So, you need to return some object from a function that the caller of the function can use further.

This is happening in your prime_counter function; you're supposed to return the l1 list from it but you're returning nothing warranting len(None) which would error out as None singleton does not have any length associated with it.

You need to return the l1 list from prime_counter :

def prime_counter(n):
    b = 1
    l1 = []
    while b <= n:
        ...
        ...
    return l1