Python函数返回null

Question

def counter(number,count):
    if (number!=1 and number%2==0):
        a=number/2
        count=count+1
        counter(a,count)
    elif (number!=1 and number%2!=0):
        a=3*(number)+1
        count=count+1
        counter(a,count)        
    else:
        print count
        return count

z=counter(13,0)
print z

count被评估为9并且确实会打印出来，但却不会返回它？在打印＆＃39; z＆＃39;

时说None

Answer 1

你忽略了递归调用;添加return语句，在counter()本身中调用counter ：

def counter(number,count):
    if (number!=1 and number%2==0):
        a=number/2
        count=count+1
        return counter(a,count)
    elif (number!=1 and number%2!=0):
        a=3*(number)+1
        count=count+1
        return counter(a,count)        
    else:
        print count
        return count

递归调用就像任何其他函数调用一样，如果你不对返回值做任何事情，它就会被丢弃。递归调用不会将其结果神奇地传递给最外层的调用。

Answer 2

你忘了返回递归的结果。

return counter(a,count)