Re: [PATCH] tags: honor COMPILED_SOURCE with apart output directory

From: Robert Jarzmik
Date: Mon Dec 12 2016 - 12:13:29 EST


Michal Marek <mmarek@xxxxxxxx> writes:

> Dne 6.12.2016 v 12:54 Robert Jarzmik napsal(a):
>> Robert Jarzmik <robert.jarzmik@xxxxxxx> writes:
>>
>>> When the kernel is compiled with an "O=" argument, the object files are
>>> not necessarily in the source tree, and more probably in another tree.
>>>
>>> In this situation, the current used check doesn't work, and
>>> COMPILED_SOURCE tags is broken with O= builds.
>>>
>>> This patch fixes it by looking for object files both in source tree and
>>> potential destination tree.
>>>
>>> Signed-off-by: Robert Jarzmik <robert.jarzmik@xxxxxxx>
>>
>> Hi Marek, ping about this patch ?
>
> Sorry, I missed the patch.
>
>>> --- a/scripts/tags.sh
>>> +++ b/scripts/tags.sh
>>> @@ -106,7 +106,9 @@ all_compiled_sources()
>>> case "$i" in
>>> *.[cS])
>>> j=${i/\.[cS]/\.o}
>>> - if [ -e $j ]; then
>>> + k=${i/"$tree"/"$O"\/}
>>> + k=${k/\.[cS]/\.o}
>>> + if [ -e $j -o -e "$k" ]; then
>
> Please use the KBUILD_OUTPUT variable and not O.
Well, I can't as far as I know. I tried, see below ...

Explanation :
- if I add just below the line "j=${i/\.[cS]/\.o}" the following :
k="$O/${j#$tree}"
echo "tree=$tree; O=$O; KBUILD_OUTPUT=$KBUILD_OUTPUT; i=$i, j=$j, k=$k"
- and I launch:
make O=out cscope COMPILED_SOURCE=1

I get these kind of lines :
tree=../; O=/home/rj/mio_linux/kernel/out; KBUILD_OUTPUT=;
i=../kernel/sched/core.c, j=../kernel/sched/core.o, k=../kernel/sched/core.o

>From here I understand that :
- $KBUILD_OUTPUT is not usable

> should only match at the beginning of the filename (so use something
> like ${i#$tree}).
Ok, I can use then : k="$O/${j#$tree}". The subtle part is when O is empty, in
which case this returns /xxxx, which doesn't look nice, while the former
expression returned either a substituted path or the source path.

> Last, but not least, the .[cS] -> .o substitution only
> needs to be done once. The k variable can use the value of j instead of i.
Yeah, definitely.

Cheers.

--
Robert