Flask(__name__)这个参数到底起了什么作用？

你可以阅读源码 flask/app.py 里面的Flask类

这只是个应用程序的名字罢了：

.. admonition:: About the First Parameter
The idea of the first parameter is to give Flask an idea of what
belongs to your application. This name is used to find resources
on the filesystem, can be used by extensions to improve debugging
information and a lot more.
So it's important what you provide there. If you are using a single
module, `__name__` is always the correct value. If you however are
using a package, it's usually recommended to hardcode the name of
your package there.
For example if your application is defined in :file:`yourapplication/app.py`
you should create it with one of the two versions below::
app = Flask('yourapplication')
app = Flask(__name__.split('.')[0])
Why is that? The application will work even with `__name__`, thanks
to how resources are looked up. However it will make debugging more
painful. Certain extensions can make assumptions based on the
import name of your application. For example the Flask-SQLAlchemy
extension will look for the code in your application that triggered
an SQL query in debug mode. If the import name is not properly set
up, that debugging information is lost. (For example it would only
pick up SQL queries in `yourapplication.app` and not
`yourapplication.views.frontend`)

__name__黑科技主要是下面这个：

@locked_cached_property
def name(self):
"""The name of the application. This is usually the import name
with the difference that it's guessed from the run file if the
import name is main. This name is used as a display name when
Flask needs the name of the application. It can be set and overridden
to change the value.
.. versionadded:: 0.8
"""
if self.import_name == '__main__':
fn = getattr(sys.modules['__main__'], '__file__', None)
if fn is None:
return '__main__'
return os.path.splitext(os.path.basename(fn))[0]
return self.import_name

@niseceric 就是看了This name is used to find resources on the filesystem
但没发现到底怎么use的

def name(self)这个是返回其name吧，注释没看懂，只看懂as a display name了。
求教