On 9 Nov 2024, at 11:42, Sebastian Feld via Cygwin <[email protected]> wrote:
>
> On Wed, Nov 6, 2024 at 4:30 PM Dimitry Andric
> <[email protected]> wrote:
>>
>> On 6 Nov 2024, at 16:11, Sebastian Feld via Cygwin <[email protected]> wrote:
>>>
>>> How can I test in a /usr/bin/make Makefile whether I am running on a
>>> 32bit or 64bit Cygwin (not Windows kernel)?
>>
>> Usually in a shell you use "uname -m", or with bash specifically, the
>> $HOSTTYPE variable.
>>
>> However in GNU make you should be able to use the internal variable
>> $(MAKE_HOST). On my Cygwin installation, this gives the value
>> "x86_64-pc-cygwin". Since I do not have any 32-bit Cygwin installation, I
>> cannot tell you what the exact value is, but my guess would be
>> "i386-pc-cygwin", or something like that.
>
> I need something which works with bsdmake (/usr/bin/bmake in Cygwin)
> and GNU /usr/bin/make. Moreover, how would a test statement for this
> look like in the Makefile? bmake doesn't have ifeq ()
BSD make has a different style for conditionals, using ".if", ".else", and so
on. These use a more familiar logical operator syntax than GNU make.
It also has a built-in variable MACHINE corresponding to "uname -m" output. So
with bmake you can do:
.if ${MACHINE} == "x86_64"
.info Go here
.else
.info Go there
.endif
However, supporting both GNU make and BSD make in one Makefile is not really
possible unless you use the lowest common denominator syntax, which is just
variable assignments and basic rules.
Otherwise, you would have to write two files, one called GNUmakefile for GNU
make, and one called Makefile for BSD make.
-Dimitry
--
Problem reports: https://cygwin.com/problems.html
FAQ: https://cygwin.com/faq/
Documentation: https://cygwin.com/docs.html
Unsubscribe info: https://cygwin.com/ml/#unsubscribe-simple
