Decorator

1. Open Closed Principle

In the case of the source code does not change , add some extra features

It is open for extension, closed for modification

1.1 principle of openness: additional new features

1.2 closed principle: Do not change the source code

2 decorator

满足开放封闭原则,在不改变原函数代码及调用方式的前提下,增加新功能

# 基本装饰器
def wrapper(func):
    def inner(*args):  # 这里的* 是聚合
        ret = func(*args)  # 这里的* 是打散
        return ret
    return inner
@wrapper
def func1(*args):
    ...

Arguments decorator

## 当装饰器装饰两个以上验证方式又不相同的函数时,就用到了带参数的装饰器
def outer(choose):  
    def wrapper(func):
        def inner(*args,**kwargs):
            if choose =='func1':
                print ('func1的判断条件')
                ret = func()
                return ret
            if choose=='func2':
                print ('func2的判断条件')
                ret = func()
                return ret
        return inner
    return wrapper
@outer('func1')   ## 这里的 outer('func1') 先执行  返回 wrapper 函数,所以就只比标准函数多一步操作
def func1():
    print('我是被装饰的函数1')
@outer('func2')    
def func2():
    
    print ('我是被装饰的函数2')
## 当然这样写很蠢;如果有操作的话可以直接对choose这个变量操作,不需要判断,因为相应的choose是不一样的

Decorative a two decorative function

def wrapper1(func):
    def inner1(*args):
        print ('1')
        func()
        print ('2')
    return inner1

def wrapper2(func):
    def inner2(*args):
        print (3)
        func()
        print (4)
    return inner2
@wrapper2         ## 这里的wrapper2 相当于把wrapper1也装饰了,相当于装饰
@wrapper1         ## inner1 = wrapper1(func1)  
def func1():      ## inner2 = wrapper2(inner1)  
    print ('new king')                  
## 最后的结果  3,1,new king 2 4

Recursion

Recursion is not a simple calls itself , he is a kind of simplify the idea , please look at the macro issues , the details to the computer

Recursive to meet two requirements

1 recursive condition: You can continue to call itself, narrowing the scope of the problem to be solved

2 Baseline conditions: export, when the conditions are not satisfied recursion

# 阶乘问题
def jc(n):
    if n==1:
        return 1
    else:
        return n*jc(n-1)

## 斐波那契数列 第n项的值
def fbnq(n):
    if n==1:
        return 1
    if n==0:
        return 0
    return fbnq(n-1)+fbnq(n-2)

Decorators & recursion