Re: [PATCH] kselftest: replace $(RM) with rm -f command

[Mathieu Desnoyers]

----- On Oct 3, 2015, at 1:55 PM, Josh Triplett josh@xxxxxxxxxxxxxxxx wrote:

> On Sat, Oct 03, 2015 at 02:11:57PM +0000, Mathieu Desnoyers wrote:
>> ----- On Oct 3, 2015, at 12:38 AM, dvhart dvhart@xxxxxxxxxxxxx wrote:
>> 
>> > On Mon, Sep 28, 2015 at 03:16:53AM +0000, Mathieu Desnoyers wrote:
>> >> ----- On Sep 27, 2015, at 10:10 PM, Wang Long long.wanglong@xxxxxxxxxx wrote:
>> >> 
>> >> > Some test's Makefile using "$(RM)" while the other's
>> >> > using "rm -f". It is better to use one of them in all
>> >> > tests.
>> >> 
>> >> I agree that this disparity appears to be unwanted. We
>> >> should settle on one or the other.
>> >> 
>> >> > 
>> >> > "rm -f" is better, because it is less magic, and everyone
>> >> > konws what is does.
>> >> 
>> >> "$(RM)" is clearly defined as a Makefile implicit variable
>> >> which defaults to "rm -f".
>> >> Ref. https://www.gnu.org/software/make/manual/html_node/Implicit-Variables.html
>> >> 
>> >> Leaving it as a variable is more flexible because then the
>> >> default behavior can be overridden if need be, which is
>> >> not the case of a hardcoded "rm -f".
>> >> 
>> >> Following your line of argumentation, we should then
>> >> invoke "gcc" directly in every Makefile because it is
>> >> less magic than "$(CC)". This makes no sense.
>> > 
>> > I don't think they can be compared so simply. Specifying a compiler is a common
>> > use case. Customizing the rm command is not, in my experience anyway, and like
>> > Michael, I would definately have to look up what RM means.
>> > 
>> > That said, I care more about consistency than which is used. Both are valid, but
>> > $(RM), while more flexible, will cost more people time to look up what it does
>> > as it isn't commonly used than any benefit we're likely to see from its use.
>> > 
>> > Meh. :-)
>> 
>> An example is "grm" when you install the opencsw repository
>> packages on Solaris. In the unlikely example where someone
>> would have a Solaris machine to build Linux, overriding
>> various command names, including "rm", can be useful. This
>> is just one example, there are probably others.
> 
> Does Solaris rm not support -f?

Yes, it does. I was merely showing this as an example where
it can be useful to override the command name, although I don't
expect anyone to have to use "grm" rather than "rm" on that
specific platform.

Thanks,

Mathieu

-- 
Mathieu Desnoyers
EfficiOS Inc.
http://www.efficios.com
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