python - 函数变为 "unbound method"。为什么？

Question

我想创建一个python类来封装一些全局变量:

class TestEnvironment(object):
    _on_out = None
    # ...

    @staticmethod
    def onOut():
        return TestEnvironment._on_out

    @staticmethod
    def setOnOut(on_out):
        TestEnvironment._on_out = on_out

# -------------------------------------------------------------------- #

def log(msg, process = None):
    print msg

# -------------------------------------------------------------------- #

if __name__ == '__main__':
    TestEnvironment.setOnOut(log)
    print log
    print TestEnvironment.onOut()    
    TestEnvironment.onOut()("A test")

运行时我得到:

<function log at 0x7fd7738122a8>
<unbound method TestEnvironment.log>
...
TypeError: unbound method log() must be called with TestEnvironment instance as first argument (got str instance instead)

似乎当我将 log 设置为 TestEnvironment 时，它变成了未绑定(bind)的方法。

我怎样才能超越它？

Answer 1

编辑:

删除了我的答案，因为它不再相关(问题似乎已经改变)。无论哪种方式 - 在 Python 3.6 上运行代码似乎都可以:

Python 3.6.1 (default, Dec 2015, 13:05:11)
[GCC 4.8.2] on linux

<function log at 0x7f2787e0a488>
<function log at 0x7f2787e0a488>
A test

这是让它在 Python 2.7 中工作的方法:

def onOut():
    return TestEnvironment.__dict__['_on_out']

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