打印出斐波纳契系列 [英] Printing out the fibonacci series

问题描述

我想写一个简单的Python程序。它应该返回一个返回连续斐波纳契数字的闭包：

I am trying to write a simple Python program. It's supposed to return a a closure that returns successive fibonacci numbers:

def fibGen():
    n_1 = 0
    n_2 = 0 
    n = 1
    def fib():
        if n_1 ==0 and n_2 ==0:
            n_1 = 1
            return n
        else:
            n = n_1 + n_2
            n_2 = n_1
            n_1 = n  
            return n
    return fib

f = fibGen()
for i in range(0,10):
    print(f())

我在运行时遇到此错误：
UnboundLocalError：在赋值之前引用的局部变量'n_1'

I get this error at run time: UnboundLocalError: local variable 'n_1' referenced before assignment

编辑：在我原来的帖子中，我没有在 fibGen 的定义中包含 n = 1 实际上是一个错字。我仍然会得到相同的错误。

In my original post, I had not included n = 1 in the definition of fibGen but it was actually a typo. I would still get the same error anyway.

推荐答案

Python在编译时确定变量的范围，基于<行为。如果您指定一个名称，或者使用它作为 import 目标（以及其他几种方式），您将绑定范围中的名称。

Python determines the scope of variables at compile time, based on binding behaviour. If you assign to a name, or use it as an import target (and a few other ways) you are binding the name in a scope.

您在 fib中绑定到 n_1 和 n_2 （） function;两者都被分配到。这使得 fib（）中的两个名称​​ local ，Python甚至不会查看周围的范围。

You are binding to n_1 and n_2 in the fib() function; both are being assigned to. This makes those two names local in fib(), and Python won't even look at the surrounding scope.

您需要覆盖此行为，您可以使用 nonlocal 语句：

You'll need to override this behaviour, and you can do this by using the nonlocal statement:

def fibGen():
    n_1 = 0
    n_2 = 0 
    def fib():
        nonlocal n_1, n_2
        if n_1 ==0 and n_2 ==0:
            n_1 = 1
            return n
        else:
            n = n_1 + n_2
            n_2 = n_1
            n_1 = n  
            return n
    return fib

非本地告诉编译器明确地，您不希望它查看绑定行为，而是将名称作为闭包。

nonlocal tells the compiler explicitly that you don't want it to look at binding behaviour but instead treat the names as closures.

接下来，您使用<$ c在测试的第一个分支中的$ c> n ，但是没有在 else 分支。您应该只传回 1 ：

Next, you are using n in the first branch of the if test, but you haven't defined it anywhere outside of the else branch. You should just return 1 there anyway:

def fibGen():
    n_1 = 0
    n_2 = 0 
    def fib():
        nonlocal n_1, n_2
        if n_1 ==0 and n_2 ==0:
            n_1 = 1
            return n_1
        else:
            n = n_1 + n_2
            n_2 = n_1
            n_1 = n  
            return n
    return fib

最后但并非最不重要的是，你可以通过使用元组赋值来交换两个变量，不需要中介：

Last but not least, you can swap two variables by using tuple assignment, no intermediaries needed:

def fibGen():
    n_1 = 0
    n_2 = 0 
    def fib():
        nonlocal n_1, n_2
        if n_1 ==0 and n_2 ==0:
            n_1 = 1
        else:
            n_1, n_2 = n_1 + n_2, n_1
        return n_1
    return fib

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