提交 8b61f748 编写于 作者: M Masahiro Yamada

gen_compile_commands: remove the warning about too few .cmd files

This warning was useful when users previously needed to manually
build the kernel and run this script.

Now you can simply do 'make compile_commands.json', which updates
all the necessary build artifacts and automatically creates the
compilation database. There is no more worry for a mistake like
"Oh, I forgot to build the kernel".

Now, this warning is rather annoying.

You can create compile_commands.json for an external module:

  $ make M=/path/to/your/external/module compile_commands.json

Then, this warning is displayed since there are usually less than
300 files in a single module.
Signed-off-by: NMasahiro Yamada <masahiroy@kernel.org>
Reviewed-by: NNick Desaulniers <ndesaulniers@google.com>
上级 3d32285f
......@@ -21,11 +21,6 @@ _FILENAME_PATTERN = r'^\..*\.cmd$'
_LINE_PATTERN = r'^cmd_[^ ]*\.o := (.* )([^ ]*\.c)$'
_VALID_LOG_LEVELS = ['DEBUG', 'INFO', 'WARNING', 'ERROR', 'CRITICAL']
# A kernel build generally has over 2000 entries in its compile_commands.json
# database. If this code finds 300 or fewer, then warn the user that they might
# not have all the .cmd files, and they might need to compile the kernel.
_LOW_COUNT_THRESHOLD = 300
def parse_arguments():
"""Sets up and parses command-line arguments.
......@@ -236,11 +231,6 @@ def main():
with open(output, 'wt') as f:
json.dump(compile_commands, f, indent=2, sort_keys=True)
count = len(compile_commands)
if count < _LOW_COUNT_THRESHOLD:
logging.warning(
'Found %s entries. Have you compiled the kernel?', count)
if __name__ == '__main__':
main()
Markdown is supported
0% .
You are about to add 0 people to the discussion. Proceed with caution.
先完成此消息的编辑!
想要评论请 注册