How do I write a "selective" Makefile?

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noob question here.

I have a directory with a lot of .c files, they're basicely libc functions that I code myself as an exercice.

I write a little main() in these files to test the functions, and I want to write a Makefile that allow me to compile only the file I want to test, for exemple:

make memset.c

And get only the executable of the code wrote in memset.c.

I tried to do something like this:

CC = gcc
CFLAGS = -Wall -Wextra -pedantic

all : %.o
    $(CC) $(CFLAGS) $<
%.o : %.c
    $(CC) $(CFLAGS) $< -c -o $@

But obviously it doens't work. I don't what to put in place of the "all". I know it's very basic, but I didn't manage to do it, and I did research but didn't find an answer to this specific question.

Thanks in advance for your help.

If you do make -n -p you get a dump of all of the built-in rules in make. In GNU Make 4.1, this includes:

%: %.o
#  recipe to execute (built-in):
        $(LINK.o) $^ $(LOADLIBES) $(LDLIBS) -o $@

So you might just needs a % in your makefile where you currently have all.

You also might find that you don't need those rules which are already built in. Suppose you have three C files, each with a main() as you specify: abs.c, div.c and fmax.c. Your Makefile needs to be no more than two lines:

CFLAGS = -Wall -Wextra -pedantic
all: abs div fmax

which would then allow you to do make abs to make the abs executable, and make all to make them all.

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You can define static pattern rules to build the object files and the executables and then invoke make with the name of the executable you want as the goal:

CC     = gcc
CFLAGS = -Wall -Wextra -pedantic

SRC := $(wildcard *.c)
OBJ := $(patsubst %.c,%.o,$(SRC))
EXE := $(patsubst %.c,%,$(SRC))

.PHONY: all obj

all: $(EXE)

obj: $(OBJ)

$(EXE): %: %.o
    $(CC) $(LDFLAGS) $< -o $@

$(OBJ): %.o: %.c
    $(CC) $(CFLAGS) $< -c -o $@

.PHONY: clean

    rm -f $(OBJ) $(EXE)


$ make memset.o

builds only memset.o,

$ make memset

builds only memset (and memset.o if needed),

$ make obj

builds all object files,

$ make # or make all

builds all executables (and object files if needed), and

$ make clean

deletes all executables and object files.

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With wildcard, you can achieve what you want.

Note that if each program depends on only one .c file, you don't need %.o rules:

CC = gcc
CFLAGS = -Wall -Wextra -pedantic

SRC  := $(wildcard *.c)
EXEC = $(SRC:%.c=%)

all: $(EXEC)

%: %.c
       $(CC) $(CFLAGS) $< -o $@ $(LDFLAGS)

And just invoke this way for instance:

make memset

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You already have most you to compile the executable selectively:

CC = gcc
CFLAGS = -Wall -Wextra -pedantic

%.o : %.c
    $(CC) $(CFLAGS) $< -c -o $@
% : %.o
    $(CC) $(LDLAGS) $< -o $@

Then you just need to call make with the target you want, the executable:

make select

If you have several sets of executable with different flags, you can use:

EX0 = drink clean
${EXE0}: % : %.o
        $(CC) $(LDLAGS) -lwater $< -o $@
EX1 = burn melt
{EX1}: % : %.o
        $(CC) $(LDLAGS) -lfire  $< -o $@

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  • The makefile is fine. The argument to make should be the resulting file you want, not the source file, so make memset.o.
  • The poster doesn't want to build a .o, they want to build an executable. If you want to create an executable from a same-named .c file you don't have to write any rules: make already has built-in rules that will do it. Just use make memset to build the memset executable from the memset.c file. Remember that you should specify targets on the make command line (that is, the things you want to be built), not sources. Make always starts with the target and works back to the sources, not the other way around.