1. 传参数

1. 向函数传递元祖

现有一个函数（三个形参），元祖t1/t2

In [2]: def fun(x,y,z):
   ...:     return x+y+z
   ...: 

In [3]: t1 = (8,9)

In [4]: t2 = (1,2,3)

• 直接给函数传入元祖,都报错：只给了一个实参

In [5]: fun(t1)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-5-2bbba3eac7d2> in <module>()
----> 1 fun(t1)

TypeError: fun() takes exactly 3 arguments (1 given)

In [6]: fun(t2)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-6-be9a8abf3158> in <module>()
----> 1 fun(t2)

TypeError: fun() takes exactly 3 arguments (1 given)

• 元祖前加一个*，t2输出结果，t1因只有2个元素，报错

In [7]: fun(*t2)
Out[7]: 6

In [8]: fun(*t1)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-8-fe16dde56c20> in <module>()
----> 1 fun(*t1)

TypeError: fun() takes exactly 3 arguments (2 given)

• 额外再给一个参数,t1满足3个参数了，但是t2多了1个参数

In [9]: fun(5,*t1)
Out[9]: 22

In [10]: fun(5,*t2)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-10-3a341823cc01> in <module>()
----> 1 fun(5,*t2)

TypeError: fun() takes exactly 3 arguments (4 given)

• 给定参数放在元祖后面，报错

In [11]: fun(*t1,5)
  File "<ipython-input-11-63abfbf858fe>", line 1
    fun(*t1,5)
SyntaxError: only named arguments may follow *expression

2.传递字典

In [1]: def fun(x,y,z):
   ...:     return x+y+z
   ...: 

In [2]: dic = {'x':1,'y':3,'z':5}

In [3]: fun(**dic)    #用**传递字典
Out[3]: 9

• 给字典换一个参数,y换成c，报错。因为c不是函数形参。但是顺序可以换。

In [4]: dic = {'x':1,'y':3,'c':5}

In [5]: fun(**dic)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-5-f4f83e494122> in <module>()
----> 1 fun(**dic)

TypeError: fun() got an unexpected keyword argument 'c'

2. 冗余参数

定义：def fun(x,y,*args,**kwargs)
args 和 kwargs 是约定俗成的冗余参数命名。

In [7]: def fun (x,*args,**kwargs):
   ...:     print x
   ...:     print args
   ...:     print kwargs
   ...:     

In [8]: fun()
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-8-69e6a439c52d> in <module>()
----> 1 fun()

TypeError: fun() takes at least 1 argument (0 given)
#因为事先已经定义了一个形参

In [9]: fun(1,2,3)
1
(2, 3)
{}

In [10]: fun(1,2,3,[1,2])
1
(2, 3, [1, 2])
{}

这说明：第一个实参对应函数定义的形参，之后的几个都对应args

In [11]: fun(1,2,3,[1,2],a=100)
1
(2, 3, [1, 2])
{'a': 100}

kwargs需要a=100这样赋值，给kwargs多传几个参数：

In [2]: def fun (x,*args,**kwargs):
    print x
    print args
    print kwargs
   ...:     

In [3]: fun(1,2,3,[1,2],a=100,'y'=5)
  File "<ipython-input-3-b0f139033a2d>", line 1
    fun(1,2,3,[1,2],a=100,'y'=5)
SyntaxError: keyword can't be an expression


In [4]: fun(1,2,3,[1,2],a=100,y=5)
1
(2, 3, [1, 2])
{'a': 100, 'y': 5}

• 更进一步：既然args输出的是元祖，那传参可以用*元祖。kwargs输出的是字典，传参可以用**字典：

In [4]: fun(1,2,3,[1,2],a=100,y=5)
1
(2, 3, [1, 2])
{'a': 100, 'y': 5}

In [5]: fun(1,2,3,[1,2],a=100,y=5,*(8,8),**{'o':4,'p':9})
1
(2, 3, [1, 2], 8, 8)
{'y': 5, 'p': 9, 'a': 100, 'o': 4}

3. 递归调用

在没有用递归调用前，举一个代码例子，计算1到100的和：

def factorial(n):
    sum = 0
    for i in range(1,n+1): #1到n
        sum += i    
    return sum
print factorial(100)

结果：

C:\Users\chawn\PycharmProjects\pyex\venv\Scripts\python.exe C:/Users/chawn/PycharmProjects/pyex/180105/cs.py
5050

Process finished with exit code 0

• 递归调用比喻：假定f(3)=3*2*1,f(2)=2*1,f(1)=1*1.f(0)=1.那f(3)=3*f(2),f(2)=2*f(1),f(1)=1*f(0)
• 递归必须有最后的默认结果，比如上面的f(0)
• 递归参数必须向默认结果收敛的: factorial(n-1)
  求和函数可改写为：

#!/usr/bin/python
# -*- coding:utf8 -*-
# author: chawn
# date:
def factorial(n):
    if n == 0:
        return 0    #f(0)=0
    else:
        return n + factorial(n-1)
print factorial(100)