Question

我必须在python（2.6）的嵌套方法中访问exec中的自由变量

def outerF():
    la = 5
    def innerF():
        str1 = "print(la)"
        exec (str1) in globals(),locals()
    innerF()
outerF()

预期结果为5，但出现以下错误

NameError：未定义名称'la'

Answer 1

对于python 2.6

def outerF():
    la={'y': 5} # exec take argu as dict and in 2.6 nonlocal is not working, 
    # have to create dict element and then process it 
    def innerF():

        str1 = "print({})".format(la['y'])
        exec (str1) in globals(),locals()
    innerF()
outerF()

Answer 2

使用非本地变量来访问嵌套函数中外部函数的变量，如下所示： Python Global, Local and Nonlocal variables

def outerF():
    la = 5
    def innerF():
        nonlocal la    #it let the la variable to be accessed in inner functions.
        str1 = "print(la)"
        exec (str1) in globals(),locals()
    innerF()
outerF()

Answer 3

请尝试对变量使用非本地语句。像这样：

def outerF():
    la = 5
    def innerF():
        nonlocal la
        str1 = "print(la)"
        exec (str1) in globals(),locals()
    innerF()
outerF()

更多信息可以在这里找到：Python Global, Local and Nonlocal variables

如何允许exec在python的嵌套方法中访问自由变量？

3 个答案: