ast.parse的filename参数有什么用？

Question

文档中写道：

ast.parse(source, filename='<unknown>', mode='exec')

    Equivalent to compile(source, filename, mode, ast.PyCF_ONLY_AST).


compile(source, filename, mode[, flags[, dont_inherit]])

    The filename argument should give the file from which the code was read;
    pass some recognizable value if it wasn’t read from a file
    ('<string>' is commonly used).

但它并没有告诉我如何从 AST 节点取回这个文件名。或者如何使用此文件名参数。它只是一个存根吗？

Answer 1

它设置co_filename代码对象的属性，该属性用于在回溯中显示文件名。除此之外，你传递什么值并不重要。

>>> c = compile('raise Exception("spam")', 'eggs', 'exec')
>>> eval(c)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "eggs", line 1, in <module>
Exception: spam