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Re: Different compilation based on filename

From: CHEN Cheng
Subject: Re: Different compilation based on filename
Date: Thu, 10 Sep 2009 21:17:55 +0800
User-agent: Mutt/1.5.17 (2007-11-01)

On Thu, Sep 10, 2009 at 02:50:47PM +0200, Christian Rogsch wrote:
> I have the following problem with a makefile: I want to compile files with 
> different compiler flags, based on the name of the file.
> Example1:
> filename is (I call it type1 file): func.f90
> make should do: $(FCOMPL) -c $(FFLAGS) $(FOPENMPFLAGS) $<
> Example2:
> filename is (I call it type2 file): smvv.f90
> make should do: $(FCOMPL) -c $(FFLAGS) $<
> Based on the makefile I have it should be (pseudocode):
> .f90.o:
> if filename = func.f90
> or filename = part.f90
> or filename = divg.f90
>       $(FCOMPL) -c $(FFLAGS) $(FOPENMPFLAGS) $<
> else
>       $(FCOMPL) -c $(FFLAGS) $<
> thus I want to create a list with different filenames (or if-then-else) 
> where make should check if this file is a type1 file and then it should 
> compile like example1, if it is not a typ1 file, it should compile like 
> example2, like shown above.

Target-specific variables may help on this problem, please see the code

# if the target is bb.o, then append more flags to $(CFLAGS)
bb.o: CFLAGS += -g


all: aa.o bb.o cc.o
    @echo ok


%.o: %.c
    @echo gcc -o $@ $*.c $(CFLAGS)


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