如何在不打印返回值的情况下正确编写返回 function 方法？

Question

def function(n):

if n % 2 != 0:
    print('weird')
elif n % 2 == 0 and n in range(2, 5):
    print('not weird')
elif n % 2 == 0 and n in range(6, 20):
    print('weird')
elif n % 2 == 0 and n > 20:
    print('not weird')
return n

while True:
    n = int(input('enter the number: '))

    print(function(n))

OUTPUT：

enter the number: 4
not weird
4

上面的代码，我不想再打印返回号4，不代表返回号怎么写？

Answer 1

将print(function(n))更改为function(n)

def function(n):
    if n % 2 != 0:
        print('weird')
    elif n % 2 == 0 and n in range(2, 5):
        print('not weird')
    elif n % 2 == 0 and n in range(6, 20):
        print('weird')
    elif n % 2 == 0 and n > 20:
        print('not weird')
    return n
while True:
    n = int(input('enter the number: '))

    function(n)

Answer 2

我的意思是你可以像这样在最后删除打印功能

def function(n):

if n % 2 != 0:
    print('weird')
elif n % 2 == 0 and n in range(2, 5):
    print('not weird')
elif n % 2 == 0 and n in range(6, 20):
    print('weird')
elif n % 2 == 0 and n > 20:
    print('not weird')
return n

while True:
    n = int(input('enter the number: '))
    function(n)

Answer 3

您也可以只从 function 返回而不打印。

Answer 4

you can use it this way.. remove the return from the function statement and also remove the print when calling this function.. If you only remove the return but still call the function using print(func(n)) then the output would be就像是

enter the number: 4
not weird
None

也许这就是你要找的？

def function(n):
if n % 2 != 0:
    print('weird')
elif n % 2 == 0 and n in range(2, 5):
    print('not weird')
elif n % 2 == 0 and n in range(6, 20):
    print('weird')
elif n % 2 == 0 and n > 20:
    print('not weird')

while True:
    n = int(input('enter the number: '))
    function(n)