Re: [SOLVED] Re: cygwin's [g]make 3.81 seems not to understand conditionals

[Carlo Florendo <subscribermail at gmail dot com>]

Carlo Florendo wrote:
> David Rothenberger wrote:
> > On 7/18/2007 9:03 PM, Carlo Florendo wrote:
> > > I've got (and attached) this simple Makefile that checks the build 
> > > platform of a machine.
> > >
> > > Here's the cut-and-pasted content of the Makefile:
> > >
> > > PLATFORM=$(shell uname -s | cut -f 1 -d " " | cut -f 1 -d "_");
> >
> > Try removing the trailing semicolon. It is not removed by make, so 
> > PLATFORM ends up being "CYGWIN;".
>
> Bingo.  That did it!
>
> So does that mean that make 3.80 was wrong then?

Please ignore this, sorry.  Both have the correct behaviour.

When expanded under Linux, the make statements become

ifeq ("Linux;", "CYGWIN)
    echo cygwin;
else
    echo linux;
endif

it will "correctly" output "linux"  but it got to the else conditional only 
because "Linux;" != "CYGWIN.

The trailing semi-colon confused me.  I was treating the Makefile as a bash 
script.


Thanks, David!

Best Regards,

Carlo

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Carlo Florendo
Softare Engineer/Network Co-Administrator
Astra Philippines Inc.
UP-Ayala Technopark, Diliman 1101, Quezon City
Philippines
http://www.astra.ph

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