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问题描述

def decorator(fn):
    def wrapper(*args, **kwargs):
        print 'With sour cream and chives!',
        return fn(*args, **kwargs)
    return wrapper

class Potato(object):
    def __call__(self):
        print 'Potato @ {} called'.format(id(self))

spud = Potato()
fancy_spud = decorator(Potato())

使用此代码,我们有两个可调用类的实例,一个实例经过修饰,一个实例很简单:

With this code we have two instances of callable class, one is decorated and one is plain:

>>> spud()
Potato @ 140408136280592 called
>>> fancy_spud()
With sour cream and chives! Potato @ 140408134310864 called

我想知道是否支持使用 @对于一个实例,可调用对象上的decorator 语法-与装饰适用于每个实例的类/方法相反。根据,@ syntax只是用于以下目的的糖:

I wonder if it is somehow supported to use the @decorator syntax on a callable for just one instance - as opposed to decorating the class / method, which would apply to every instance. According to this popular answer, the @syntax is just sugar for:

function = decorator(function)

但这是过度简化?经过我所有的尝试,似乎只有在语法出现在 def class ,空白或 @another_decorator

But is it an over-simplification? With all my half-baked attempts it seems only to work when the syntax occurs before def, class, whitespace or @another_decorator.

@decorator
baked = Potato()

这是 SyntaxError

baked = Potato()
@decorator
baked

SyntaxError

@decorator
def baked(_spud=Potato()):
    return _spud()

有效,但是很丑陋,有点作弊。

Works, but is ugly and kinda cheating.

推荐答案

您提出的问题是:

function = decorator(function)

但是,更准确地说是

@decorator
def function():
    pass

是语法糖用于:

def function():
    pass
function = decorator(function)

装饰器专门用于装饰函数,方法或类 definitions 。引入了类修饰符的描述了语法:

Decorators are designed to decorate function, method or class definitions, specifically. The PEP that introduced class decorators describes the grammar:

decorated: decorators (classdef | funcdef)

funcdef: 'def' NAME parameters ['->' test] ':' suite

如您所见,装饰器必须紧接 classdef funcdef ,因此无法直接在可调用类的实例上使用它。

As you can see, a decorator must come immediately before a classdef or funcdef, so there is no way to use it directly on an instance of a callable class.

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08-20 11:16