代码如下:
#coding=utf8__author__ = ‘administrator’
# 当函数的参数不确定时，可以使用*args和**kwargs。*args没有key值，**kwargs有key值
def fun_var_args(farg, *args): print ‘args:’, farg for value in args: print ‘another arg:’,value
# *args可以当作可容纳多个变量组成的list或tuplefun_var_args(1, ‘two’, 3, none)
#args: 1#another arg: two#another arg: 3#another arg: none
def fun_var_kwargs(farg, **kwargs): print ‘args:’,farg for key in kwargs: print ‘another keyword arg:%s:%s’ % (key, kwargs[key])
# myarg1,myarg2和myarg3被视为key， 感觉**kwargs可以当作容纳多个key和value的dictionaryfun_var_kwargs(1, myarg1=’two’, myarg2=3, myarg3=none)# 输出：#args: 1#another keyword arg:myarg1:two#another keyword arg:myarg2:3#another keyword arg:myarg3:none
def fun_args(arg1, arg2, arg3): print ‘arg1:’, arg1 print ‘arg2:’, arg2 print ‘arg3:’, arg3
myargs = [‘1’, ‘two’, none] # 定义列表fun_args(*myargs)
# 输出：#arg1: 1#arg2: two#arg3: none
mykwargs = {‘arg1’: ‘1’, ‘arg2’: ‘two’, ‘arg3’: none} # 定义字典类型fun_args(**mykwargs)
# 输出：#arg1: 1#arg2: two#arg3: none
# 两者都有def fun_args_kwargs(*args, **kwargs): print ‘args:’, args print ‘kwargs:’, kwargs
args = [1, 2, 3, 4]kwargs = {‘name’: ‘beginman’, ‘age’: 22}fun_args_kwargs(args,kwargs)# args: ([1, 2, 3, 4], {‘age’: 22, ‘name’: ‘beginman’})# kwargs: {}
fun_args_kwargs(1,2,3,a=100)#args: (1, 2, 3)#kwargs: {‘a’: 100}
fun_args_kwargs(*(1,2,3,4),**{‘a’:none})#args: (1, 2, 3, 4)#kwargs: {‘a’: none}