On Sat, Oct 03, 2015 at 02:11:57PM +0000, Mathieu Desnoyers wrote:
> ----- On Oct 3, 2015, at 12:38 AM, dvhart dvhart@infradead.org 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@huawei.com 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?