makefile CFLAGS ignore $(mysql_config --libs)? - mysql

Question

I make a toy makefile example to test mysql, but the makefile does not recognize mysql_config.
this is the makefile script:
CFLAGS = -g -O2 -Wall -Wextra -Isrc -rdynamic $(OPTFLAGS)
LDLIBS = $(OPTLIBS)
SOURCES =$(wildcard *.c)
OBJECTS = asd
all: LDLIBS += $(mysql_config --libs_r) -lm
CFLAGS += -Isrc $(mysql_config --cflags)
all: $(OBJECTS)
When i run make all, it only execute:
cc -g -O2 -Wall -Wextra -Isrc -rdynamic -Isrc asd.c -lm -o asd
Where did all the mysql CFLAGS and LDLIBS go? Or is there something wrong with my script?
this returns when i type 'mysql_config --cflags' in the shell, for demonstration:
-I/usr/include/mysql

Answer 1

The content $(mysql_config --libs_r) is intended to ask the shell to invoke that command and replace the string with its output.
But, make uses the $(...) syntax to expand variables. So, your attempt at running a shell command mysql_config --libs_r is actually being interpreted as expanding a make variable named mysql_config --libs_r, of which there is not one, and so you get an empty string here.
You need to escape the $(...) syntax from make so that it's passed to the shell.
Also, your indentation seems to imply you want both LDLIBS and CFLAGS to be target-specific variables on the all target, however if that's really what you want you have to use a backslash at the end of the first line. Simply indenting the line doesn't make it a continuation of the previous line.
You want this:
all: LDLIBS += $$(mysql_config --libs_r) -lm \
CFLAGS += -Isrc $$(mysql_config --cflags)
There are some efficiency issues with this as it will run mysql_config twice for every compile and link operation. Much more efficient would be something like:
mysql_LIBS := $(shell mysql_config --libs_r)
mysql_FLAGS := $(shell mysql_config --cflags)
then use the make variables $(mysql_LIBS) and $(mysql_FLAGS)