老齐python-基础9(函数)

>>> def foo(x,y,z,*args,**kargs):
... print(x)
... print(y)
... print(z)
... print(args)
... print(kargs)
...
>>> foo('qiwsir',2,"python") #多参数各种类型调用适用各种场景
qiwsir
2
python
()
{}
>>> foo(1,2,3,4,5)
1
2
3
(4, 5)
{}
>>> foo(1,2,3,4,5,name="qiwsir")
1
2
3
(4, 5)
{'name': 'qiwsir'}

>>> def bar():
... print("I am in bar()")
...
>>> def foo(func): #使用func传递函数
... func()
...
>>> foo(bar)
I am in bar()

def power_seq(func,seq):
return [func(i) for i in seq]def pingfang(x):
return str(x) #return x ** 2求平方if __name__ == "__main__":
num_seq = [111,3.14,2.91]
r = power_seq(pingfang, num_seq)
print(num_seq)
print(r)

def foo():
def bar():
print("bar() is running")
bar()
print("foo() is running")print(foo())

>>> def foo():
... a = 1
... def bar():
... b = a + 1
... print("b=",b)
... bar()
... print("a=",a)
...
>>> foo()
b= 2
a= 1
>>> def foo(): #另外一种定义方法就会出错Python解析器认定该变量应在bar()内部建立，而不是引用外部对象，所以报错
... a = 1
... def bar():
... a = a + 1
... print("bar()a=",a)
... bar()
... print("foo()a=",a)
...
>>> foo()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 6, in foo
File "<stdin>", line 4, in bar
UnboundLocalError: local variable 'a' referenced before assignment

>>> def foo():
... a = 1
... def bar():
... nonlocal a
... a = a + 1
... print("bar()a=",a)
... bar()
... print("foo()a=",a)
...
>>> foo()
bar()a= 2
foo()a= 2

#!/usr/bin/env python3
# encoding: utf-8"""
@version: ??
@author: tajzhang
@license: Apache Licence
@file: zhongli.py
@time: 2018/2/28 15:50
"""def weight(g):
def cal_mg(m):
return m * g
return cal_mgw = weight(10) #函数对象多层引用，动态函数对象，闭包
mg = w(10)
print(mg)g0 = 9.78046
w0 = weight(g0)
mg0 = w0(10)
print(mg0)