python - Python 的 locals() 有什么作用？

Question

Python 的 locals()文档说:

Update and return a dictionary representing the current local symbol table. Free variables are returned by locals() when it is called in function blocks, but not in class blocks.

1. 当前本地符号表中到底有哪些内容？

2. 在函数最开始调用locals()是否保证局部符号表与参数dict相同？

   例如，如果我们有以下程序:

   def foo(x, y):
       print(locals() == {'x': 1, 'y': 2})
   foo(1, 2)
   

   无论我们使用什么平台和 Python 实现，它总是输出 True 吗？

Answer 1

What are in the current local symbol table exactly?

嗯，你可以自己看看。在您的 Python 程序的最开始，locals() 返回的符号表如下所示:

print(locals())

哪些输出:

{'__doc__': None, '__spec__': None, '__package__': None, '__builtins__': 
<module 'builtins' (built-in)>, '__name__': '__main__', '__loader__': 
<class '_frozen_importlib.BuiltinImporter'>, 
'__file__': 'C:\\Users\\$Name$\\Desktop\\script.py'}

符号表由一些“魔法变量”和一些关于当前 Python 文件的信息组成。比如__file__键，里面包含了你当前源文件的名字。 locals 返回的内容的描述与符号表的定义非常匹配:

In computer science, a symbol table is a data structure used by a language translator such as a compiler or interpreter, where each identifier in a program's source code is associated with information relating to its declaration or appearance in the source. - Wikipedia: Symbol table

_{(强调我的)}

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Is the local symbol table guaranteed to be the same as the argument dict if locals() is called at the very beginning of a function?

答案是¹是。函数有自己的作用域。而且，正如名称所暗示的那样，locals() 仅返回当前范围的本地标识符。因此，在函数内部对 locals() 的调用不能被程序的外部范围更改。例如。

>>> var = 10 # global variable
>>> locals()['var'] # var is accessible in the current scope
10
>>> def func():
    print(locals()['var']) # but not in this scope. Python will raise an error


>>> func()
Traceback (most recent call last):
  File "<pyshell#17>", line 1, in <module>
    func()
  File "<pyshell#16>", line 2, in func
    print(locals()['var'])
KeyError: 'var'
>>> 

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¹<子>很难完全理解你在第二个问题中的要求，所以如果我的回答不相关，我深表歉意。但我相信你会问:如果我在函数的开头调用 locals()，在我的定义中，locals 返回的 dict() 是否保证保留一样吗？。如果不是这种情况，请更新您的问题，我会尝试重新回答。