blocks|key|651500|text|若要修改函数内的全局变量，必须使用global关键字。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|651501|当您尝试在不使用行的情况下执行此操作时|651502|global+counter|code-block|syntax|javascript|651503|在增量的定义中，创建了一个名为counter的局部变量，以防止弄乱整个程序可能依赖的计数器变量。|651504|请注意，您只需要在修改变量时使用global；您可以从增量中读取计数器，而不需要使用global语句。|651505|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

To modify a global variable inside a function, you must use the global keyword.

When you try to do this without the line

<pre><code>global counter
</code></pre>

inside of the definition of increment, a local variable named counter is created so as to keep you from mucking up the counter variable that the whole program may depend on.

Note that you only need to use global when you are modifying the variable; you could read counter from within increment without the need for the global statement.

blocks|key|651605|text|您需要使用global+statement，以便修改全局变量计数器，而不是局部变量：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|651606|counter+=+0

def+increment():
++global+counter
++counter+%2B=+1

increment()|code-block|syntax|javascript|651607|如果定义counter的封闭作用域不是全局作用域，那么在Python3.x上可以使用nonlocal+statement。在Python2.x上，在同样的情况下，您将无法重新分配非本地名称counter，因此您需要将counter设置为可变名称并对其进行修改：|style|CODE|651608|counter+=+[0]

def+increment():
++counter[0]+%2B=+1

increment()
print+counter[0]++#+prints+'1'|651609|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/py3k/reference/simple_stmts.html#the-global-statement|1|http://docs.python.org/py3k/reference/simple_stmts.html#the-nonlocal-statement^0|5|G|0|0|0|4|7|2M|7|30|7|16|I|1|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|13|8|@$A|14|B|15|K|L]|$A|16|B|17|K|L]|$A|18|B|19|K|L]]|9|@$A|1A|B|1B|1|1C]]|C|$]]|$1|M|3|N|5|F|7|1D|8|@]|9|@]|C|$G|H]]|$1|O|3|-4|5|6|7|1E|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

You need to use the <a href="http://docs.python.org/py3k/reference/simple_stmts.html#the-global-statement">global statement</a> so that you are modifying the global variable counter, instead of a local variable:

<pre><code>counter = 0

def increment():
 global counter
 counter += 1

increment()
</code></pre>

If the enclosing scope that <code>counter</code> is defined in is not the global scope, on Python 3.x you could use the <a href="http://docs.python.org/py3k/reference/simple_stmts.html#the-nonlocal-statement">nonlocal statement</a>. In the same situation on Python 2.x you would have no way to reassign to the nonlocal name <code>counter</code>, so you would need to make <code>counter</code> mutable and modify it:

<pre><code>counter = [0]

def increment():
 counter[0] += 1

increment()
print counter[0] # prints '1'
</code></pre>

blocks|key|437262|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|437263|counter+=+0

def+increment():
++global+counter
++counter+%2B=+1

increment()|code-block|syntax|javascript|437264|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

try this

<pre><code>counter = 0

def increment():
 global counter
 counter += 1

increment()
</code></pre>

blocks|key|437313|text|Python不是纯粹的词法作用域。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|437314|看看这个：Using+global+variables+in+a+function|offset|length|437315|这个是：https://www.saltycrane.com/blog/2008/01/python-variable-scope-notes/|437316|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/423379/using-global-variables-in-a-function-other-than-the-one-that-created-them/|1|https://www.saltycrane.com/blog/2008/01/python-variable-scope-notes/^0|0|5|10|0|0|4|1W|1|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@$D|T|E|U|1|V]]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@$D|X|E|Y|1|Z]]|A|$]]|$1|H|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]|P|$5|K|L|M|A|$N|Q]]]]

Python is not purely lexically scoped.

See this: <a href="https://stackoverflow.com/questions/423379/using-global-variables-in-a-function-other-than-the-one-that-created-them/">Using global variables in a function</a>

and this: <a href="https://www.saltycrane.com/blog/2008/01/python-variable-scope-notes/" rel="nofollow noreferrer">https://www.saltycrane.com/blog/2008/01/python-variable-scope-notes/</a>

blocks|key|651806|text|为了回答主题行中的问题，*是的，Python中有闭包，只是它们只适用于函数内部，而且(在Python2.x中)闭包是只读的；您不能将名称重新绑定到不同的对象(但如果对象是可变的，则可以修改其内容)。在Python3.x中，可以使用nonlocal关键字修改闭包变量。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|651807|def+incrementer():
++++counter+=+0
++++def+increment():
++++++++nonlocal+counter
++++++++counter+%2B=+1
++++++++return+counter
++++return+increment

increment+=+incrementer()

increment()+++#+1
increment()+++#+2|code-block|syntax|javascript|651808|*这个问题最初是关于Python中的闭包。|651809|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/reference/simple_stmts.html?highlight=nonlocal#nonlocal^0|37|8|37|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

To answer the question in your subject line,* yes, there are closures in Python, except they only apply inside a function, and also (in Python 2.x) they are read-only; you can't re-bind the name to a different object (though if the object is mutable, you can modify its contents). In Python 3.x, you can use the <a href="https://docs.python.org/3/reference/simple_stmts.html?highlight=nonlocal#nonlocal" rel="noreferrer"><code>nonlocal</code></a> keyword to modify a closure variable.

<pre><code>def incrementer():
 counter = 0
 def increment():
 nonlocal counter
 counter += 1
 return counter
 return increment

increment = incrementer()

increment() # 1
increment() # 2
</code></pre>

* The question origially asked about closures in Python.

blocks|key|651903|text|代码抛出UnboundLocalError的原因已经在其他答案中得到了很好的解释。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|651904|但在我看来，您正在尝试构建像itertools.count()一样工作的东西。|651905|所以你为什么不尝试一下，看看它是否适合你的情况：|651906|>>>+from+itertools+import+count
>>>+counter+=+count(0)
>>>+counter
count(0)
>>>+next(counter)
0
>>>+counter
count(1)
>>>+next(counter)
1
>>>+counter
count(2)|code-block|syntax|javascript|651907|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/itertools.html#itertools.count^0|4|H|0|E|H|E|H|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]]|E|$]]|$1|H|3|I|5|6|7|15|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|16|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

The reason of why your code throws an <code>UnboundLocalError</code> is already well explained in other answers.

But it seems to me that you're trying to build something that works like <a href="http://docs.python.org/library/itertools.html#itertools.count" rel="noreferrer"><code>itertools.count()</code></a>. 

So why don't you try it out, and see if it suits your case:

<pre><code>&gt;&gt;&gt; from itertools import count
&gt;&gt;&gt; counter = count(0)
&gt;&gt;&gt; counter
count(0)
&gt;&gt;&gt; next(counter)
0
&gt;&gt;&gt; counter
count(1)
&gt;&gt;&gt; next(counter)
1
&gt;&gt;&gt; counter
count(2)
</code></pre>

What am I doing wrong here?

<pre><code>counter = 0

def increment():
 counter += 1

increment()
</code></pre>

The above code throws an <code>UnboundLocalError</code>.

Don't understand why UnboundLocalError occurs (closure)

Python

我在这里做错了什么？counter = 0def increment():  counter += 1increment()上面的代码抛出了一个UnboundLocalError。

问不理解为什么会发生UnboundLocalError (闭包)
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