Detect missing "early return"?

Python Help
1

I am reading code which looks like this:

def foo():
    if ...:
        ...
        ...
        ...

I prefer this:

def foo():
    if not ...:
        return
    ...
    ...
    ...

In above example I used three “…” lines. As soon as there are more then 5 lines I would like to use “early return”.

Is there any tool in the Python galaxy which does this:

  • detects a missing “early return”
  • and/or updates the code to use “early return”

Related: My guidelines: use early return

1 Like
2

You could use the inspect module, e.g.:

src=inspect.getsource(foo)
if src[src.find('if'):src.find('return')].count('\n') > 5:
  ...
1 Like
3

Check to see if flake8 has a plugin for your desired check, and if not, you could write it yourself.

1 Like
4

Sorry, I fail to understand how your can work reliably. There can be several “if” and several “return” in the source code.