Force Makefile to execute script before building targets

user_247265 0
Viewed 23412

I am using Makefiles.

However, there is a command (zsh script) I want executed before any targets is executed. How do I do this?

Thanks!

target makefile
4 Answers

There are several techniques to have code executed before targets are built. Which one you should choose depends a little on exactly what you want to do, and why you want to do it. (What does the zsh script do? Why do you have to execute it?)

You can either do like @John suggests; placing the zsh script as the first dependency. You should then mark the zsh target as .PHONY unless it actually generates a file named zsh.

Another solution (in GNU make, at least) is to invoke the $(shell ...) function as part of a variable assignment:

ZSH_RESULT:=$(shell zsh myscript.zsh)

This will execute the script as soon as the makefile is parsed, and before any targets are executed. It will also execute the script if you invoke the makefile recursively.

user_130510

Just make that a dependancy of one of the other targets

foo.obj : zsh foo.c 
   rule for compileing foo.c

zsh: 
   rule for running zsh script.

or alternatively, make your first target depend on it

goal: zsh foo.exe
user_2348150

There is a solution without modifying your existing Makefile (main difference with the accepted answer). Just create a makefile containing:

.PHONY: all

all:
    pre-script
    @$(MAKE) -f Makefile --no-print-directory $(MAKECMDGOALS) MAKE='$(MAKE) -f Makefile'
    post-script

$(MAKECMDGOALS): all ;

The only drawback is that the pre- and post- scripts will always be run, even if there is nothing else to do. But they will not be run if you invoke make with one of the --dry-run options (other difference with the accepted answer).

user_17737980
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