递归代码返回 None

问题描述

我真的不明白，为什么要代码

I really do not understand, why the code

def isIn(char, aStr):
    ms = len(aStr)/2
    if aStr[ms] == char:
        print 'i am here now'
        return True
    elif char>aStr[ms] and not ms == len(aStr)-1:
        aStr = aStr[ms+1:]
    elif char <aStr[ms] and not ms == 0:
        aStr = aStr[0:ms]
    else:
        return False
    isIn(char, aStr)

print isIn('a', 'ab')

继续返回 None.它打印我现在在这里"，但它不会返回 True，正如下一行所说.为什么?

does keep on returning None. it prints 'i am here now', but it does not return True, just as the next line says. Why?

推荐答案

你可能想要在最后一行的 return :

You probably want a return on the last line:

return isIn(char, aStr)

如果没有它，函数在没有看到 return 的情况下终止时只会返回 None.

Without it, the function simply returns None when it terminates without seeing a return.

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