On Thu, Oct 11, 2007 at 10:52:03AM +0900, Charles Plessy wrote: > Dear mentors, > > in a package I prepare, there is the following line in a source/Makefile: > > CPPFLAGS=-O3 -funroll-loops -march=i686 -mfpmath=sse -msse -mmmx CPPFLAGS is for the C PreProcessor. So it's supposed to have things like -Iinclude -DFOO=BAR. CFLAGS is for the C compiler itself, so it's supposed to have things like -O3 -funroll-loops -std=gnu99 -Wall -Wextra. But it only matters if you're using the implicit rules and the binaries are built from multiple source files or are otherwise compiled and linked in separate invocations. BTW there's LDFLAGS too for linker options like -Wl,-soname,libfoo.so.1 -Wl,-O2 (here it's assumed that LD=gcc thus the -Wl, thing).
I think that Debian packages shouldn't have subarch-specific options on any arch, since the same binaries may/(have to be able to) be run on variation on that arch. In the case of i386, the binaries have to be able to run on a real 386 [0]. I think all the arch options here will (maybe) cause the binary to fail on such a machine. Justin [0] My understanding is that the packaged kernels don't support 386 but with a software emulation of some math instruction patched in, 386 is advertised as being supported with binary packages. -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of "unsubscribe". Trouble? Contact [EMAIL PROTECTED]
