為什么答案是(2,0)不是(2,1)
def sub_sum(num): ? ?sum1=0 ? ?sum2=0 ? ?if num%2==0: ? ? ? ?sum1=num+sum1 ? ?else: ? ? ? ?sum2=num+sum2 ? ?return sum1,sum2 L=[1,2] for i in L: ? ?result=sub_sum(i) print(result)?
def sub_sum(num): ? ?sum1=0 ? ?sum2=0 ? ?if num%2==0: ? ? ? ?sum1=num+sum1 ? ?else: ? ? ? ?sum2=num+sum2 ? ?return sum1,sum2 L=[1,2] for i in L: ? ?result=sub_sum(i) print(result)?
2021-02-03
舉報
2021-09-17
你函數(shù)里面不寫for循環(huán)的嘛,判斷是否為偶數(shù)的前面加個list的for循環(huán)就好了
2021-02-18
sum2 = 0?
2021-02-16
因為每次調(diào)用sub_sum(num)函數(shù)時,sum1和sum2都會重新被定義一次