c1 = new_counter(10) c2 = new_counter(20) print c1(), c2(), c1(), c2() outputs ： 11 21 12 22 def new_counter(n): class Counter(object): def __init__(self, num): self.num =num def __call__(self): self.num += 1 return self.num return Counter(n) 结束以后才想出来的，觉得比较麻烦，还有更简洁的方法吗？
谢谢~
def new_counter(num): count = num def counter(): nonlocal ount count += 1 return count return counter def new_counter(num): count = [num] def counter(): count[0] += 1 return count[0] return counter  | 1 jellyshen Jan 20, 2016 ```python def new_count(n): print(n) def f(): print(n+1) return f ``` 函数 f 里不能影响变量 n 的状态，还请大神们指教。 |
 | 2 deangl Jan 20, 2016 def new_counter(init_val): _counter = [init_val, ] def add1(): _counter[0] += 1 return _counter[0] return add1 |
 | 3 zhangchioulin Jan 20, 2016 LZ 是面试的什么? |
| 4 deangl Jan 20, 2016 ``` def new_counter(init_val): _counter = [init_val, ] def add1(): _counter[0] += 1 return _counter[0] return add1 ``` |
| 5 deangl Jan 20, 2016 啊，求教怎么发贴带格式，怎么删自己的贴？ |
| 6 jellyshen Jan 20, 2016 问个问题，怎么在 V2EX 的回复里面排版代码？ ```python def new_count(n): print(n) def f(): print(n+1) return f ``` |
| 7 zhangchioulin Jan 20, 2016 @deangl 这个...哥你用的什么语言?为什么都 return 了还能执行? |
 | 8 lxy42 Jan 20, 2016 不会是搜狐吧 |
| 9 lxy42 Jan 20, 2016  1 用闭包就行了 |
| 10 deangl Jan 20, 2016 1 @zhangchioulin 我打不出缩进。。。。 这个意思： def new_counter(init_val): ...._counter = [init_val, ] ....def add1(): ........_counter[0] += 1 ........return _counter[0] ....return add1 |
 | 11 tkpc Jan 20, 2016 5 new_counter = lambda n : (x+n for x in xrange(sys.maxint)).next |
 | 12 fghzpqm Jan 20, 2016 via iPad import itertools new_counter = itertools.count |
 | 13 swirling Jan 20, 2016 via iPhone high order function ，这时候你就和面试官聊聊函数式什么就好了。 |
 | 14 dapang1221 Jan 20, 2016 python 生成器？用 yield 传出来就好了，不要用 return ……但是调用起来又得用 next 了。。。 def counter(n): ....while True: ........n=(yield n+1) |
 | 15 snailsir Jan 20, 2016 via iPhone 1 闭包 |
 | 16 mengzhuo Jan 20, 2016 1 我来上个优雅点的吧： def new_counter(i): ....def c(): ........c.var += 1 ........return c.var ....c.var = i ....return c c10 = new_counter(10) c20 = new_counter(20) print c10(), c20(), c10(), c20() #11, 21, 12, 22 |
 | 17 airqj Jan 20, 2016 via Android 记不清实现了 应该是闭包 |
 | 18 ChiangDi Jan 21, 2016 via Android 这个是最简单的闭包应用啊， Python 不熟，随便用 js 码个 function new_count(n) { return function() { n ＋＋ console.log(n) } 还有比这个更简单的实现吗？ |
 | 19 elvba Jan 21, 2016 function new_counter($t) { return function () use (&$t) { return $t += 1; }; } |
 | 20 SpicyCat Jan 21, 2016 1 |
 | 21 master13 Jan 21, 2016 1 感觉做一个生成器，或者 yield 把…… |
 | 22 cppgohan Jan 21, 2016 用 itertools 偷懒可以吗? ``` def new_counter(c): counter = itertools.count(c) def f(): return counter.next() return f; ``` ``` class new_counter(itertools.count): def __init__(self, *args, **kwargs): super(new_counter, self).__init__(*args, **kwargs) def __call__(self): return self.next() ``` |
| 23 cppgohan Jan 21, 2016 1 |
 | 24 ttycode Jan 21, 2016 via Android 闭包，或者匿名函数 |
 | 25 motopig Jan 21, 2016 ``` php function counter($num) { return function () use (&$num) { return ++$num."<br>"; }; } $c1 = counter(10); $c2 = counter(20); echo $c1(); echo $c2(); echo $c1(); echo $c2(); ``` |
 | 26 youkochan Jan 21, 2016 1 def new_counter(_num): def foo(): num = _num while True: num += 1 yield num a = foo() return a.next c1 = new_counter(10) c2 = new_counter(20) print c1(), c2(), c1(), c2() |
 | 27 shajiquan Jan 21, 2016  1 我觉得楼主提出的 class 方式就挺好的，清晰直观，不用理会变量的作用域带来的困扰。#16 楼的方法挺好玩的。 |
 | 28 qiu0130 Jan 21, 2016 C pa pa 重载()运算符. |
 | 29 necomancer Jan 22, 2016 2 ```python3 def new_counter(m): ....def counter(m): ........while 1: ............m +=1 ............yield m ....return counter(m).__next__ c1 = new_counter(10) c2 = new_counter(20) print(c1(), c2(), c1(), c2()) ``` 没 16 楼的简洁…… |
| 30 necomancer Jan 22, 2016 啊……原来 26 楼已经是这个办法了…… |
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