我对这种范围行为感到困惑：

class Bar:
    def __init__(self):
        for fn in ["open","openW","remove","mkdir","exists","isdir","listdir"]:
            print "register", fn
            def func_wrapper(filename):
                print "called func wrapper", fn, filename
            setattr(self, fn, func_wrapper)

bar = Bar()
bar.open("a")
bar.remove("b")
bar.listdir("c")

这给出了输出：

register open
register openW
register remove
register mkdir
register exists
register isdir
register listdir
called func wrapper listdir a
called func wrapper listdir b
called func wrapper listdir c

但是我希望这func_wrapper始终是正确的功能。我知道范围func_wrapper是整个函数，但是我在每次循环迭代中都重新定义了它，最后一个实例保存在属性中。我还尝试在的func_wrapper = None下面添加内容，setattr但这无济于事（也想知道我...）。

我是盲人吗？我什至没有真正看到如何解决/解决此问题。

解决方案

要么与

class Bar:
    def __init__(self):
        for fn in ["open","openW","remove","mkdir","exists","isdir","listdir"]:
            print "register", fn
            def func_wrapper(filename, fn=fn):
                print "called func wrapper", fn, filename
            setattr(self, fn, func_wrapper)

或者更强大地

def mkwrapper(fn):
    def func_wrapper(filename):
        print "called func wrapper", fn, filename
    func_wrapper.__name__ = fn
    return func_wrapper

class Bar:
    def __init__(self):
        for fn in ["open","openW","remove","mkdir","exists","isdir","listdir"]:
            print "register", fn
            func_wrapper = mkwrapper(fn)
            setattr(self, fn, func_wrapper)

在您的原始示例中，所有生成的函数都访问相同的外部变量fn，该变量在每次循环运行中都会更改。在更正的示例中，这是可以避免的。