无法访问通过更改函数内部的locals()创建的变量 [英] Can't access variable created by altering locals() inside function

问题描述

在调用以变量名作为参数的函数时，我需要分配一些变量.

I need to assign some variables when calling a funcion with the name of the variable as an argument.

因此，我遍历具有所需名称的元组，并通过 locals()字典对其进行分配.

Therefore, I loop through a tuple with the desired names, assigning them via the locals() dict.

它可以工作，但是后来我无法按名称访问它们-即使在函数本身内部.

It works, but then I can't access them by name - even inside the function itself.

def function():
    var_names = ("foo","bar","foobar")
    for var_name in var_names:
        locals()[var_name] = len(var_name)
    print foo

投掷:

Traceback (most recent call last):
  File "error_test.py", line 8, in <module>
    function()
  File "error_test.py", line 5, in function
    print foo
NameError: global name 'foo' is not defined

使用以下代码，效果很好:

With the following code it works well:

def function():
    var_names = ("foo","bar","foobar")
    for var_name in var_names:
        locals()[var_name] = len(var_name)
    print locals()["foo"]

locals()字典是否仅包含普通的函数变量?为什么不起作用?

Isn't it that the locals() dict contains just the normal function variables? Why isn't it working?

推荐答案

编写时:

for var_name in var_names:
    locals()[var_name] = len(var_name)

您修改 locals()字典:

作为@ user2357112 在文档中适当链接:

as @user2357112 aptly linked in the docs:

注意

此字典的内容不应修改；更改可能不会影响解释器使用的局部变量和自由变量的值.

The contents of this dictionary should not be modified; changes may not affect the values of local and free variables used by the interpreter.

因此修改 locals() 会删除局部变量，因此出现 NameError: global name 'foo' is not defined 错误.

So modifying locals() deletes the local variables, hence the NameError: global name 'foo' is not defined error.

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