Re: Makefile with .mod files

[Erik Schnetter <schnetter at cct dot lsu dot edu>]

On Nov 3, 2007, at 14:11:03, Ignacio Fernández Galván wrote:

> --- FX Coudert <fxcoudert@gmail.com> wrote:
>
> > This is a change from 4.2, IIRC, and it was done because users
> > complained that .mod files had their timestamps updated while not
> > changing, which triggered massive useless recompilations because of
> > Makefile dependencies ;-)
>
> Fair enough, I think that's OK
>
> > > So, is there a solution for this?
> >
> > Yes, I believe there is; the following Makefile seems to work fine:
> >
> > all: a.o b.o
> >
> > a.o: a.f90 b.mod
> >          gfortran -c $<
> >
> > b.o: b.f90
> >          gfortran -c $<
> >
> > b.mod: b.f90 b.o
> >          @true
> >
> > a.mod: a.f90 a.o
> >          @true
>
> It works... until one deletes the .mod files (which can happen by
> accident), then make doesn't know how to build them. I guess I could
> use something like "test -e $@ || gfortran -c $<" instead of  
> "@true"...
> Anyway, there is another (minor) problem: with that solution make  
> still
> thinks there is something to build and does not inform that everything
> is up to date, even though it doesn't actually build anything.

Since .mod files and .o files are generated at the same time, I  
prefer the following:

all: a.o b.o

a.o: a.f90 b.o
	gfortran -c $<

b.o: f.90
	gfortran -c $<

Of course, in reality you use rules, so that would read:

all: a.o b.o

%.o: %.f90
	gfortran -c $*.f90

a.o: b.o

Finally, if you use GNU make you can also write:

a.o a.mod: a.f90
	gfortran -c $<

to indicate that both files are created.  (At least I think this is  
possible; maybe it is only possible with rules instead of explicit  
targets, or there is another gotcha which I'm missing at the moment.   
As I said above, I don't usually mention .mod files in Makefiles.)

-erik

--
Erik Schnetter <schnetter@cct.lsu.edu>

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