你好C 2018-01-27
如果在函数内部需要修改全局变量那么需要使用global关键字
x=1
def mfun():
global x
x=2
print(x)
>>> mfun()
2内部函数的的作用域在外部函数作用于之内,及只能在外部函数内调用内部函数
def outside():
print("正在调用outside")
def inside():
print("正在调用inside")
inside()
outside()
inside()#这句话是错的
正在调用outside
正在调用inside
Traceback (most recent call last):
File "C:\Users\ENVY\Desktop\learnning Python\text.py", line 7, in <module>
inside()
NameError: name 'inside' is not defineddef line_conf():
def line(x):
return 2*x+1
return line # return a function object
my_line = line_conf()
print(my_line(5))def line_conf():
b = 15
def line(x):
return 2*x+b
return line # return a function object
b = 5
my_line = line_conf()
print(my_line(5)) # 返回25在内部函数中只能对外部函数的局部变量进行访问,但是不能修改,如果需要修改则需要用到nonlocal关键字,委屈求全可以使用“容器类型”代替
def line_conf():
b = 15
def line(x):
nonlocal b
b=20
return 2*x+b
return line # return a function object
b = 5
my_line = line_conf()
print(my_line(5)) #返回30def line_conf():
b = [15]
def line(x):
b[0]=20
return 2*x+b[0]
return line # return a function object
my_line = line_conf()
print(my_line(5)) #返回30