Re: Conditionals, step one

[Radly]

Derek Clegg wrote:
> 
> On Sep 16, 2010, at 5:56 AM, Radly wrote:
> 
>> Problems like this make me feel like a beginner...
>> 
>> I have a project that needs to detect OSTYPE and set LIBS and LIBDIRS
>> accordingly, but I can't get basic conditionals working.  This doesn't
>> work
>> on either Mac OSX or Ubuntu Linux; they both use Make 3.81.  Here's a
>> stripped-down demo of my problem:
>> 
>> OS = $OSTYPE
>> 
>> ifeq ($(value OS), darwin10.0)
>>  TXT = GOT_IT
>> else
>>  TXT = MISSED_IT
>> endif
>> 
>> all:
>>      @echo OS = $(value OS)
>>      @echo $(TXT)
>> 
>> And here's the printed output:
>> OS = darwin10.0
>> MISSED_IT
>> 
>> I've tried multiple variations on the ifeq test, and they all fail,
>> producing the "MISSED_IT" text.
>> 
>> Surely I'm missing something that will be obvious to someone else.  What
>> might it be?
> 
> Use $(info ...) to print the value of your variables. You'll see that OS
> is equal to STYPE, not what you intended. Do this instead:
>   OS = $(OSTYPE).
> 
> Derek
> 
> 

Thanks, that was helpful, except that 'info' seems to be a no-op.  After
trying a lot of variations and simplifications, I finally got the following
to almost work:

ifeq (${value OSTYPE},linux-gnu)
  TXT = GOT_IT
else
  TXT = MISSED_IT
endif

all:
        @echo ${TXT}

Now I would like to know how to pick up OSTYPE directly from the shell,
without having to type "make OSTYPE = $OSTYPE".  If I enter it that way, the
makefile prints out "GOT_IT", but if I just type "make", it prints
"MISSED_IT".

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