为什么在带有块的 Python 中不可调用可调用？

Question

关于为什么在 Python 3.5with语句中不可调用的可调用对象的任何押韵或原因？

class Seriously(object):
    def __init__(self, name):
        self.name = name

    def __enter__(self):
        print("Enter " + self.name)

    def __call__(self):
        print("Call " + self.name)

    def __exit__(self, type, value, traceback):
        print("Exit " + self.name)

a = Seriously('a')
a.__enter__()
a()
a.__enter__()

with Seriously('b') as b:
    b()

严重产生

Enter a
Call a
Enter a
Enter b
Exit b
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-6-91a5d11e1b2e> in <module>()
     18 
     19 with Seriously('b') as b:
---> 20     b()

TypeError: 'NoneType' object is not callable

我在PEP 343 中缺少什么？

Answer 1

您不会从该__enter__方法返回任何内容，该方法由 python 编译器转换为None. 您应该从 中返回要受as子句约束的值__enter__。

该with声明被删除为

mgr = (EXPR)
exit = type(mgr).__exit__ 
value = type(mgr).__enter__(mgr)  ### <--- usually the same as mgr.__enter__()
exc = True
try:
    try:
        VAR = value               ### <--- value returned by __enter__, VAR from with/as statement
        BLOCK
    except:
        exc = False
        if not exit(mgr, *sys.exc_info()):
                raise
finally:
    if exc:
        exit(mgr, None, None, None)

（来自您在问题中链接的 PEP）。