Re: [PATCH v2 2/5] Makefile: add debug target

[Todor Minchev <todor.minchev@linux.intel.com>]

On Fri, 2017-02-10 at 10:58 -0800, Jianxun Zhang wrote:
> > On Feb 9, 2017, at 11:17 AM, Todor Minchev <todor.minchev@linux.intel.com> wrote:
> > 
> > A debug version of the rmc binary can be built by using the debug
> > Makefile target. This will include debug symbols and will disable
> > compiler optimizations.
> > 
> > Example:
> > 
> > make debug
> > 
> > Signed-off-by: Todor Minchev <todor.minchev@linux.intel.com>
> > ---
> > Makefile | 2 ++
> > 1 file changed, 2 insertions(+)
> > 
> > diff --git a/Makefile b/Makefile
> > index c58047a..fdd936f 100644
> > --- a/Makefile
> > +++ b/Makefile
> > @@ -23,6 +23,8 @@ ALL_OBJS := $(RMC_TOOL_OBJ) $(RMC_LIB_OBJ)
> > RMC_CFLAGS := -Wall -I$(TOPDIR)/inc
> > 
> > all: rmc
> > +debug: RMC_CFLAGS += -DDEBUG -g -O0
> > +debug: rmc
> > 
> I missed your feedback of V1, just recap it here also with my comment:
> 
> - "Yup, the above will have the same effect as 'make debug'.
> I thought that 'make debug' might be a convenient way to build a debug
> binary? Do you think it makes sense to keep this extra target for
> convenience since it doesn't affect the other usages of make?"
> 
> You could have to come back to CFLAGS when the a hard coded debug flags is not enough.
> 
> And, why pass -DDEBUG in RMC_CFLAGS when invoking gcc to compile rmc? I don’t see this macro DEBUG in the current rmc code.

I am planning to use the DEBUG macro in the future to add some more
verbosity to binaries built with 'make debug'

> 
> Thanks
> > $(ALL_OBJS): %.o: %.c
> > 	$(CC) -c $(CFLAGS) $(RMC_CFLAGS) $< -o $@
> > -- 
> > 2.11.1
> > 
>