Python:递归函数中的变量绑定(bind)

Question

我在 Python 中使用类似于以下的函数遇到了一些奇怪的事情:

    def foo(x):
        if int(x)!=4:
            x = raw_input("Wrong guess, please enter new value: " )
            foo(x)
        else:
            print "Good Job! %s was right"%x
        return x

    value = foo(x=raw_input("Guess a Number between 1 and 10: "))
    print value

例如，如果我输入:“1”然后“2”然后“3”然后“4”，我会打印出以下内容:

Good Job! 4 was right
2

这很令人困惑，因为函数似乎成功地识别了正确的答案，但在这样做之后它返回的值是给出的第二个响应，而不是最近的响应。

谁能解释一下这个递归函数中“x”的绑定(bind)是怎么回事？

Answer 1

让我们看看!

value = foo(raw_input())
# foo is the same as in the question, I won't repeat it here
print value

在你的 foo 中，你得到这个:

# foo(1) calls foo(2) and sets x to 2
# foo(2) calls foo(3) and sets x to 3
# foo(3) calls foo(4) and sets x to 4
# foo(4) does:
print "Good Job! 4 was right"
return 4 # to foo(3)
# foo(3) does:
return 4 # to foo(2)
# foo(2) does:
return 3 # to foo(1)
# foo(1) does:
return 2 # to main

因为 main 的返回值(从最外层的递归)是 2，所以 value 就是这个值。

要解决此问题，您可以使其迭代:

def iter_foo(x):
    while int(x) != 4:
        x = raw_input("Wrong guess. Try again! ")
    print "Good Job! %s was right" % x
    return x

或者让每个递归返回新函数的结果

def recurse_foo(x):
    if int(x) != 4:
        return foo(raw_input("Wrong guess. Try again! "))
    else:
        print "Good Job! %s was right!" % x
        return x