python - 取消注册单次派送？

Question

有没有办法为泛型“注销”已注册的函数？

例如：

from functools import singledispatch

@singledispatch
def foo(x):
    return 'default function'

foo.register(int, lambda x: 'function for int')

# later I would like to revert this.

foo.unregister(int) # does not exist - this is the functionality I am after

Answer 1

singledispatch仅用于附加；你不能真正取消注册任何东西。

但与所有 Python 一样，实现可以强制取消注册。以下函数将向unregister()singledispatch 函数添加一个方法：

def add_unregister(func):
    # build a dictionary mapping names to closure cells
    closure = dict(zip(func.register.__code__.co_freevars, 
                       func.register.__closure__))
    registry = closure['registry'].cell_contents
    dispatch_cache = closure['dispatch_cache'].cell_contents
    def unregister(cls):
        del registry[cls]
        dispatch_cache.clear()
    func.unregister = unregister
    return func

这会进入singledispatch.register()函数的闭包以访问实际的registry字典，因此我们可以删除已注册的现有类。我还清除了dispatch_cache弱引用字典以防止它介入。

您可以将其用作装饰器：

@add_unregister
@singledispatch
def foo(x):
    return 'default function'

演示：

>>> @add_unregister
... @singledispatch
... def foo(x):
...     return 'default function'
... 
>>> foo.register(int, lambda x: 'function for int')
<function <lambda> at 0x10bed6400>
>>> foo.registry
mappingproxy({<class 'object'>: <function foo at 0x10bed6510>, <class 'int'>: <function <lambda> at 0x10bed6400>})
>>> foo(1)
'function for int'
>>> foo.unregister(int)
>>> foo.registry
mappingproxy({<class 'object'>: <function foo at 0x10bed6510>})
>>> foo(1)
'default function'