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In my makefile, I am trying to capture the output from a shell function call on a make variable containing the command string without success. When I run the shell function on the string directly, it works. I don't understand what the difference is between running the shell function on the command string versus running the shell function on a make variable containing the command string. 

PG_CONFIG = "/usr/pgsql-9.4/bin/pg_config/"

PG_INCLUDE1 = $(shell $$PG_CONFIG)
PG_INCLUDE2 = $(shell /usr/pgsql-9.4/bin/pg_config --includedir-server)

.PHONY: print

print:
    @echo "PG_CONFIG="$(PG_CONFIG)
    @echo "PG_INCLUDE1="$(PG_INCLUDE1)
    @echo "PG_INCLUDE2="$(PG_INCLUDE2)

The output is:

$make -f Makefile.test print
PG_CONFIG=/usr/pgsql-9.4/bin/pg_config/
PG_INCLUDE1=
PG_INCLUDE2=/usr/pgsql-9.4/include/server
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1
  • try PG_INCLUDE1 = $(shell $(PG_CONFIG)) Commented Apr 16, 2015 at 17:46

2 Answers 2

Reset to default
3

As it stands in GNU Make Manual:

Variable and function references in recipes have identical syntax and semantics to references elsewhere in the makefile. They also have the same quoting rules: if you want a dollar sign to appear in your recipe, you must double it (‘$$’). For shells like the default shell, that use dollar signs to introduce variables, it’s important to keep clear in your mind whether the variable you want to reference is a make variable (use a single dollar sign) or a shell variable (use two dollar signs).

The $$ is replaced with a single dollar. In your example:

PG_INCLUDE1 = $(shell $$PG_CONFIG)

becomes:

PG_INCLUDE1 = $(shell $PG_CONFIG)

Next, $PG_CONFIG is executed in shell just as if you typed it in the terminal. And as there is no PG_CONFIG variable defined in the current subshell $PG_CONFIG is replaced with nothing. However, if you defined PG_CONFIG environment variable before running make it would work as you wished with $$ in Makefile:

$ export PG_CONFIG="aoeuidhtn"
$ make
PG_CONFIG=/usr/pgsql-9.4/bin/pg_config/
PG_INCLUDE1=aoeuidhtn
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5 Comments

Always use := when assigning values to variables within make files.
@mhradek: Because?
Per https://ftp.gnu.org/old-gnu/Manuals/make-3.79.1/html_chapter/make_6.html the colon tells the processor to only read and assign once.
First, link you posted describes an ancient version of make - 3.79.1 was released in 2000 ftp.gnu.org/gnu/make. Second, you said what := does but didn't explain why should it always be used?
Same applies: gnu.org/software/make/manual/make.html Perhaps always is too strong of a word. How about "Consider using simple assignment over recursive assignment whenever possible to be explicit and avoid inadvertent assignment issues".
2

I see two problems. The definition of PG_CONFIG ends with a "/" so it is being treated as a directory. And your variable reference in PG_INCLUDE1 should be $(PG_CONFIG) not $$PG_CONFIG.

Try this:

PG_CONFIG = "/usr/pgsql-9.4/bin/pg_config"

PG_INCLUDE1 = $(shell $(PG_CONFIG))
PG_INCLUDE2 = $(shell /usr/pgsql-9.4/bin/pg_config --includedir-server)

.PHONY: print

print:
    @echo "PG_CONFIG="$(PG_CONFIG)
    @echo "PG_INCLUDE1="$(PG_INCLUDE1)
    @echo "PG_INCLUDE2="$(PG_INCLUDE2)
Improve this answer

3 Comments

There's also no reason to keep the make variables outside the quoted string in the shell lines (they can even go inside a single quoted string).
Your example yields Makefile:9: *** missing separator. Stop.
Nevermind. Needed to replace the spaces before @echo with tabs.

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