Messages in this thread | | | Subject | Re: [PATCH v4] checkpatch: get default codespell dictionary path from package location | From | Joe Perches <> | Date | Mon, 11 Oct 2021 09:50:53 -0700 |
| |
On Mon, 2021-10-11 at 14:49 +0300, Peter Ujfalusi wrote: > The standard location of dictionary.txt is under codespell's package, on > my machine atm (codespell 2.1, Artix Linux): > /usr/lib/python3.9/site-packages/codespell_lib/data/dictionary.txt > > Since we enable the codespell by default for SOF I have constant: > No codespell typos will be found - \ > file '/usr/share/codespell/dictionary.txt': No such file or directory [] > diff --git a/scripts/checkpatch.pl b/scripts/checkpatch.pl [] > + # Try to find the codespell install location to use it as default path > + if (($codespell || $help) && which("codespell") ne "" && which("python") ne "") { > + my $codespell_dict = `python -c "import os.path as op; import codespell_lib; print(op.join(op.dirname(codespell_lib.__file__), 'data', 'dictionary.txt'), end='')" 2> /dev/null`; > + $codespellfile = $codespell_dict if (-e $codespell_dict); > + }
This is really hard to read.
Can this be written something like
my $python_codespell_dict = << "EOF" import os.path as op import codespell_lib codespell_dir = op.dirname(codespell_lib.__file__) codespell_file = op.join(codespell_dir, 'data', 'dictionary.txt') print(codespell_file) EOF ; my $codespell_dict = `python3 -c "$python_codespell_dict" 2> /dev/null`;
| |