编写python闭包的更好方法是什么？

Question

我的问题可以通过简单的是或否回答:

• 是的,italic's"封闭样式"是编写python闭包的更好方法 - 换句话说 - italic's"封闭符号"不会破坏并且没有任何警告

• 不,italic关闭是有问题的,bold"封闭符号"是唯一的方法

为了以防万一,如果有人想知道"这件事是什么",那么为什么"他"不使用标准封口 - 标准形式,它让我想要逃跑.

def example():

    # The standard X_x closure...
    #
    def bold(predecessor):
        def successor():
            return "<bold/>" + predecessor() + "</bold>"
        return successor

    # ...this looks sooo much better ^.^
    #
    def italic(predecessor):
        x = predecessor
        def successor():
            return "<italic/>" + x() + "</italic>"
        return successor

    def underline(predecessor):
        x = predecessor
        def successor():
            return "<underline/>" + x() + "</underline>"
        return successor

    @italic
    @bold
    @underline
    def trademark():
        return "This trademark cant be changed."

    print(trademark())
    #example() prints <italic/><bold/><underline/>This trademark cant be changed.</underline></bold</italic>

Answer 1

当然,你可以这样做:

def italic(predecessor):
    x = predecessor
    def successor():
        return "<italic/>" + x() + "</italic>"
    return successor

就像你能做的那样:

def italic(predecessor):
    x = predecessor
    x2 = x
    def successor():
        return "<italic/>" + x2() + "</italic>"
    return successor

要么

def italic(predecessor):
    x = predecessor
    x2 = x
    x3 = x2
    def successor():
        return "<italic/>" + x3() + "</italic>"
    return successor

但你为什么要这样做？

编写装饰器的最佳方法是functools.wraps:

from functools import wraps

def italic(f):
    @wraps(f)
    def decorated():
        return "<italic/>" + f() + "</italic>"
    return decorated

但如果你真的想要与众不同:

import functools

def italic(f):
    return functools.update_wrapper(lambda: "<italic/>" + f() + "</italic>", f)