Issue

I want to write a Makefile which would run tests. Test are in a directory './tests' and executable files to be tested are in the directory './bin'.

When I run the tests, they don't see the exec files, as the directory ./bin is not in the $PATH.

When I do something like this:

EXPORT PATH=bin:$PATH
make test

everything works. However I need to change the $PATH in the Makefile.

Simple Makefile content:

test all:
    PATH=bin:${PATH}
    @echo $(PATH)
    x

It prints the path correctly, however it doesn't find the file x.

When I do this manually:

$ export PATH=bin:$PATH
$ x

everything is OK then.

How could I change the $PATH in the Makefile?

Solution

Did you try export directive of Make itself (assuming that you use GNU Make)?

export PATH := bin:$(PATH)

test all:
    x

Also, there is a bug in you example:

test all:
    PATH=bin:${PATH}
    @echo $(PATH)
    x

First, the value being echoed is an expansion of PATH variable performed by Make, not the shell. If it prints the expected value then, I guess, you've set PATH variable somewhere earlier in your Makefile, or in a shell that invoked Make. To prevent such behavior you should escape dollars:

test all:
    PATH=bin:$$PATH
    @echo $$PATH
    x

Second, in any case this won't work because Make executes each line of the recipe in a separate shell. This can be changed by writing the recipe in a single line:

test all:
    export PATH=bin:$$PATH; echo $$PATH; x

Answered By - Eldar Abusalimov

Answer Checked By - Marilyn (WPSolving Volunteer)

Tuesday, February 6, 2024

[SOLVED] How I could add dir to $PATH in Makefile?

Issue

Solution