python中装饰器的装饰顺序和执行顺序

在python语言中“装饰器”模式被设计成语言内置的设计模式，使得使用该模式变得简单，仅仅需要在目标函数前使用‘@’符号加装饰器名就能达到目的。但如果同一个目标函数需要进行多次装饰，那么调用目标函数时装饰器的执行顺序是怎样的呢？

这里有个例子，有个目标函数叫lunch，中午调用一下来决定午饭吃什么

# coding: utf-8
import random

# 饭馆
restaurants = {
    'KFC': ['hamburger', 'chicken wings', 'ice cream', 'salmon fish'],
    'xiapu': ['Beef hotpot', 'Mutton hotpot', 'Tomato hotpot'],
}

def chose_food(f):
    print 'decorator: chose_food enter'
    def func(*args, **kwargs):
        print 'decorator: chose_food::func enter'
        a_rstrnt = args[0]
        food = random.choice(restaurants[a_rstrnt])
        return f(food, *args, **kwargs)

    print 'decorator: chose_food exit'
    return func

def restaurant(f):
    print 'decorator: restaurant enter'
    def func(*args, **kwargs):
        print 'decorator: restaurant::func enter'
        a_rstrnt = random.choice(restaurants.keys())
        return f(a_rstrnt, *args, **kwargs)

    print 'decorator: restaurant exit'
    return func

@restaurant
@chose_food
def lunch(food, addr, then):
    print "We're going to {} for '{}', then {}".format(addr, food, then)

lunch('have a rest')

输出结果

decorator: chose_food enter
decorator: chose_food exit
decorator: restaurant enter
decorator: restaurant exit
decorator: restaurant::func enter
decorator: chose_food::func enter
We're going to KFC for 'hamburger', then have a rest

通过打印结果可以看出装饰器装饰的顺序是“从下到上”依次执行，而执行的顺序是“从上到下”。

原因是装饰器在这个例子中起的作用是对目标函数包装，两个装饰器装饰过后，是这种情形：

lunch(then){} = deco.restaurant(then) { 
                    deco:chose_food(rstrnt, then) { 
                        lunch(food, rstrnt, then){} 
                    }
                }

lunch(then)被调用时自然按照由外到内的顺序执行。