How can I force Make to execute a recipe all the time

问题

The current Makefile has something like this:

target1 : lib1.a  lib2.a

target2 : lib1.a  lib3.a

target3 : lib3.a 

lib1.a:
    $(MAKE) -C sub_dir all

I want to change this Makefile so that wherever a target depends on lib1.a, it always run the command "$(MAKE) -C sub_dir all", always. Another words, in the above example, target1 and target2 will always run "$(MAKE) -C sub_dir all". Is there any way I can do that?

I know the following does not work:

target1 :  lib2.a
    $(MAKE) -C sub_dir all

target2 :   lib3.a
    $(MAKE) -C sub_dir all

target3 : lib3.a

Because if lib2.a has no update, the the command does not run. I have one restriction, I only control lib1.a, I cannot change how lib2.a is built.

Any help is appreciated!

UPDATE: I used the following solution:

target1 : lib1.a  lib2.a

target2 : lib1.a  lib3.a

target3 : lib3.a 

lib1.a: relay

.PHONY: relay
relay:
    $(MAKE) -C sub_dir all

The GNU Make help says that FORCE is not as efficient as .PHONY, but from your explanation, it seems to me FORCE is better. Do I mis-understand?

回答1:

I think a Force Target is what you are looking for here.

lib1.a: FORCE
        $(MAKE) -C sub_dir all

FORCE: ;

As compared to a .PHONY rule (as suggested by @MadScientist here) this will not in-and-of-itself force all the targets that depend on lib1.a to rebuild all the time. It will only do that if lib1.a actually changes.

A better solution though is likely to get rid of the sub-make entirely and actually have the dependency information for lib1.a available to this makefile (either directly or via an included makefile in sub_dir) because that avoids this entire "force this recipe because I can't tell you how to tell if the target needs updating" problem.

来源：https://stackoverflow.com/questions/26226106/how-can-i-force-make-to-execute-a-recipe-all-the-time

标签

makefile

gnu-make