linux-kernel - Re: [PATCH v4] checkpatch: get default codespell dictionary path from package location

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Date:   Mon, 11 Oct 2021 09:50:53 -0700
From:   Joe Perches <joe@...ches.com>
To:     Peter Ujfalusi <peter.ujfalusi@...ux.intel.com>, apw@...onical.com
Cc:     dwaipayanray1@...il.com, lukas.bulwahn@...il.com,
        linux-kernel@...r.kernel.org
Subject: Re: [PATCH v4] checkpatch: get default codespell dictionary path
 from package location

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`;