How I could add dir to $PATH in Makefile?

Question

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?

Answer 1

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