return之前有值，return之后返回None,为什么

Question

def search(num,seq=[]):    if len(seq)==1:        print(seq[0])        return seq[0]    elif num<seq[len(seq)//2]:         seq=seq[0:len(seq)//2]         search(num,seq=seq[:])    elif num > seq[len(seq)//2]:         seq=seq[len(seq)//2+1:]         search(num,seq=seq[:])    else:        return seq[len(seq)//2] seq=[1,2,3,4,5,6,7,8] seq.sort()print(seq)print(search(4,seq=seq[:]))[1, 2, 3, 4, 5, 6, 7, 8]4None望解决，这是一个二分法使用案例

Answer 1

每一个函数在末尾都隐藏了一句 return None,除非你写了自己的return语句

def search(num):
    global seq    if len(seq) == 1:        print(seq[0])        return seq[0]    elif num < seq[len(seq)//2]:
        seq = seq[0:len(seq)//2]        print(seq)        return search(num)    elif num > seq[len(seq)//2]:
        seq = seq[len(seq)//2+1:]        print(seq)        return search(num)    else:        return seq[len(seq)//2]


seq = [1, 2, 3, 4, 5, 6, 7, 8]
seq.sort()print(seq[:])print(search(4))