[英]Python: variable bindings in recursive functions
I came across some weirdness in Python using a function similar to the following: 我使用类似于以下的函数在Python中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
If I enter, for instance: "1" then "2" then "3" then "4", I get the following printed out: 如果我输入,例如:“1”然后“2”然后“3”然后“4”,我打印出以下内容:
Good Job! 4 was right
2
This is confusing, since it seems the function is successfully identifying the right answer, but after doing so it is returning a value that was the 2nd response given, instead of the most recent response. 这很令人困惑,因为看起来该函数似乎成功识别了正确的答案,但在这样做之后,它返回的值是给定的第二个响应,而不是最近的响应。
Can anyone explain what's going on with the binding of "x" in this recursive function? 任何人都可以解释这个递归函数中绑定“x”的情况吗?
Let's see! 让我们来看看!
value = foo(raw_input())
# foo is the same as in the question, I won't repeat it here
print value
Inside your foo, you get this: 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
Since the return value to main (from the outermost recursion) is 2
, that's what value
remains. 由于main(从最外面的递归)的返回值是2
,这就是剩下的value
。
To fix this, you can either make it iterative: 要解决这个问题,你可以让它迭代:
def iter_foo(x):
while int(x) != 4:
x = raw_input("Wrong guess. Try again! ")
print "Good Job! %s was right" % x
return x
Or make EACH recursion return the result of the new function 或者使EACH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
So it's mostly because you are missing a return
in 所以这主要是因为你错过了return
def foo(x):
if int(x)!=4:
x = raw_input("Wrong guess, please enter new value: " )
foo(x) # <-- need return
else:
print "Good Job! %s was right"%x
return x
value = foo(x=raw_input("Guess a Number between 1 and 10: "))
print value
What's happening is it calls foo(1)
, which is not equal to 4, assigns x
to 2, and that is the last value that x
gets assigned to, so 发生的事情是调用foo(1)
,它不等于4,将x
赋值为2,这是x
被赋值的最后一个值,所以
foo(x)
returns a value, but it isn't assigned to anything or returned from the recursive call. 递归foo(x)
返回一个值,但它没有分配给任何东西或从递归调用返回。 x
received was 2, so that is what the initial foo(1)
returns 的最后一个值x
接收为2,所以这就是初始foo(1)
返回 So either 所以要么
x = foo(x)
Or 要么
return foo(x)
On the line I highlighted 在线上我突出显示
Try the following code: 请尝试以下代码:
def foo(x):
if int(x)!=4:
x = raw_input("Wrong guess, please enter new value: " )
return 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
You can think each call to the function foo, creates a new variable x. 你可以认为每次调用函数foo,都会创建一个新的变量x。 Thus, the recursive calls make a sequence of variables x1, x2, x3, etc. Thus when a function call returns, your local variable x is unchanged. 因此,递归调用产生一系列变量x1,x2,x3等。因此,当函数调用返回时,您的局部变量x不变。 That's why you still get 2 instead of the assignment from the last recursive call (4). 这就是为什么你仍然得到2而不是最后一次递归调用的赋值(4)。 If you want to change the variable passed into a function, you need to pass it by reference (or object), rather than by value. 如果要更改传递给函数的变量,则需要通过引用(或对象)而不是值传递它。 [See this article for more details: http://uvesway.wordpress.com/2013/02/18/passing-arguments-to-a-function-by-value-by-reference-by-object/] [有关详细信息,请参阅此文章: http : //uvesway.wordpress.com/2013/02/18/passing-arguments-to-a-function-by-value-by-reference-by-object/]
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