Date: Fri, 04 Jan 2002 11:20:55 -0800 (PST) From: John Baldwin <jhb@FreeBSD.org> To: Stephen Montgomery-Smith <stephen@math.missouri.edu> Cc: freebsd-hackers@FreeBSD.ORG, Alfred Perlstein <bright@mu.org> Subject: Re: Tell gcc I have a i686 Message-ID: <XFMail.020104112055.jhb@FreeBSD.org> In-Reply-To: <3C35F513.2F16AC08@math.missouri.edu>
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On 04-Jan-02 Stephen Montgomery-Smith wrote:
> Alfred Perlstein wrote:
>>
>> * Stephen Montgomery-Smith <stephen@math.missouri.edu> [020104 12:02] wrote:
>> > I want to create a Makefile for a C program that includes some Pentium
>> > II specific inline assembler code. How do I tell the compiler whether
>> > we are compiling on a i686?
>> >
>> > For Linux, I can do something like this (for gnu-make)
>> > Arch = $(shell arch)
>> > cc ...... -DArch .....
>> >
>> > and inside the program
>> >
>> > #ifdef i686
>> >
>> > But arch doesn't exist on FreeBSD.
>>
>> Isn't this somewhat trivial?
>>
>> ARCH=i686
>> CFLAGS+=-D${ARCH}
>>
>> ?
>>
>
>
> What I want is a makefile that automatically detects whether it is on an
> i686 or not (not for me to tell it so).
This doesn't support a user who wants to compile an app that they want to run
on some other machine. :) On FreeBSD, you can see if CPUTYPE is set from
make.conf, but it's not required to be set.
--
John Baldwin <jhb@FreeBSD.org> <>< http://www.FreeBSD.org/~jhb/
"Power Users Use the Power to Serve!" - http://www.FreeBSD.org/
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