Lets say I have files:
Libs:
one.cpp, one.h
two.cpp, two.h
three.cpp, three.h
Program:
program.cpp
Is there way, to create Makefile which will compile only that *.cpp which were modified from last compilation?
Currently I have something like that:
SRCS = one.cpp two.cpp three.cpp
OBJS = $(SRCS:.cpp=.o)
all: $(OBJS) program
.cpp.o:
g++ -Wall -c $<
program:
g++ -Wall $(OBJS) program.cpp -o program
clean:
rm -f $(OBJS) program
I works fine, but when I compile my program and then change two.cpp or two.h I need to run "make clean" first, because when I secondly run "make" I get:
Nothing to be done for 'all'.
I would like to change my Makefile in that way, it would recognize my changes and recompile that file and its dependencies (if one.cpp uses code from two.cpp which was modified, both files should be recompiled).
So if I modify two.cpp, make should do:
g++ -Wall -c two.cpp
g++ -Wall $(OBJS) program.cpp -o program
But if one.cpp uses code from two.cpp which was modified, make shold do:
g++ -Wall -c one.cpp
g++ -Wall -c two.cpp
g++ -Wall $(OBJS) program.cpp -o program
First we make the object files prerequisites of the executable. Once this is done, Make will rebuild program whenever one of the SRCS changes, so we don't need OBJS as an explicit target:
all: program
program: $(OBJS)
g++ -Wall $(OBJS) program.cpp -o program
Then we make the header files prerequisites of the objects, so that if we change three.h, Make will rebuild three.o:
$(OBJS): %.o : %.h
And finally since one.cpp uses code from two.cpp by means of two.h (I hope), we make two.h a prerequisite of one.o:
one.o: two.h
And to make things cleaner and easier to maintain we use automatic variables:
program: $(OBJS)
g++ -Wall $^ program.cpp -o $#
Put it all together and we get:
SRCS = one.cpp two.cpp three.cpp
OBJS = $(SRCS:.cpp=.o)
all: program
$(OBJS): %.o : %.h
one.o: two.h
.cpp.o:
g++ -Wall -c $<
program: $(OBJS)
g++ -Wall $^ program.cpp -o $#
clean:
rm -f $(OBJS) program
There are a few more things we could do (like adding program.o to OBJS), but this is enough for today.
Add the files a command depends upon to run to the right of the target name.
Example:
default: hello.c
gcc -o hello.bin hello.c
install: hello.bin
cp hello.bin ../
All you need to do is tell make that the .o file depends on the .cpp file:
%.cpp.o: %.cpp
g++ -Wall -c -o $# $<
Related
I'm doing a Makefile to make objects with the same gcc command. This file looks like this:
SRCLIB = main.c srv.c
OBJLIB = main.o srv.o
CC = gcc
CCFLAGS = -Wall -Werror
$(OBJLIB) : $(SRCLIB)
$(CC) $(CCFLAGS) -c $^ -o $#
The objetive is to execute this like:
gcc -Wall -c read_line.c -o read_line.o
gcc -Wall -c client.c -o client.o
But I don't know how to do it, and everything I tested is not working. Is it even possible to do this in a Makefile?
Your makefile expands to this, after the variables are expanded:
main.o srv.o : main.c srv.c
$(CC) $(CCFLAGS) -c $^ -o $#
In make, using multiple targets in explicit rules like this is the same as writing the rule multiple times, once for each target. So, this is the same as this:
main.o : main.c srv.c
$(CC) $(CCFLAGS) -c $^ -o $#
srv.o : main.c srv.c
$(CC) $(CCFLAGS) -c $^ -o $#
This means that if either of the source files changes, BOTH object files will be recreated (since each object depends on both sources, not just their own source file).
Further, in your compile line you use the variable $^ which expands to all the prerequisites. So your compile lines will expand to:
gcc -Wall -Werror -c main.c srv.c -o main.o
gcc -Wall -Werror -c main.c srv.c -o srv.o
which is illegal: if you use -c with the -o option you can only compile one source file.
Make has built-in rules that already know how to compile files, so there's no need to write your own. You can just write this:
SRCLIB = main.c srv.c
OBJLIB = main.o srv.o
CC = gcc
CCFLAGS = -Wall -Werror
.PHONY: all
all: $(OBJLIB)
and that's all you need.
I've got a basic, minimal makefile to compile C++ source; a stripped-down version looks like this:
TARGET = main
OBJ = $(patsubst %.cpp,%.o,$(wildcard *.cpp))
CPPFLAGS = -std=c++11 -Wall
build: $(TARGET)
$(TARGET): $(OBJ)
#echo "$? --> $#"
g++ $(CPPFLAGS) -o $# $?
#echo
%.o: %.cpp
#echo "$< --> $#"
g++ $(CPPFLAGS) -c $<
#echo
When building from scratch, everything works as expected:
clock.cpp --> clock.o
g++ -std=c++11 -Wall -c clock.cpp
main.cpp --> main.o
g++ -std=c++11 -Wall -c main.cpp
clock.o main.o --> main
g++ -std=c++11 -Wall -o main clock.o main.o
After a change in one of the source files, make rebuilds the corresponding object file, but for some reason skips the existing (unchanged) object file when trying to link (clock.o --> main instead of clock.o main.o --> main):
clock.cpp --> clock.o
g++ -std=c++11 -Wall -c clock.cpp
clock.o --> main
g++ -std=c++11 -Wall -o main clock.o
When running make again, without changing any file, the linking works as expected:
clock.o main.o --> main
g++ -std=c++11 -Wall -o main clock.o main.o
What's the reason for this behavior, and how could it be prevented?
I'm running GNU Make 4.1 on i686 Linux 4.4
You use wrong automatic variable in g++ $(CPPFLAGS) -o $# $? recipe.
https://www.gnu.org/software/make/manual/html_node/Automatic-Variables.html:
$?
The names of all the prerequisites that are newer than the target,
with spaces between them.
So Make does exactly what you ask it to do: link main target from all updated prerequisites and skip untouched prerequisites.
g++ $(CPPFLAGS) -o $# $^ recipe will fix the problem.
I am having some confusion about how the '$<' and '$#' macros can be used with a list of elements. My ultimate goal is to compile a directory of C source files into executables with the same name, no extensions. Also I do not want to make object files.
Here is a simple makefile which I would like to upgrade using the macros.
CC = gcc -ansi -std=c99
CCFLAGS = -Wall -pedantic -O3
all : progA progB progC
progA : progA.c
$(CC) $(CCFLAGS) $< -o $#
progB : progB.c
$(CC) $(CCFLAGS) $< -o $#
progC : progC.c
$(CC) $(CCFLAGS) $< -o $#
This works fine, but I don't like the redundancy of the commands. I have found a workaround that works, and a near solution, but was hoping there might be a clear option.
workaround:
CC = gcc -ansi -std=c99
CCFLAGS = -Wall -pedantic -O3
PRGS := $(patsubst %.c,%,$(wildcard *.c))
all :
make $(PRGS)
% : *.c
$(CC) $(CCFLAGS) $#.c -o $#
What I don't like here is the call to make within the command. Running 'make' delivers a message in my terminal that looks something like this:
make[1]: Entering directory '/path/to/dir'
...actual commands...
make[1]: Leaving directory '/path/to/dir'
I am assuming this has to do with opening the same makefile and the [1] refers to the second file descriptor in the open file table (or something along those lines).
near solution:
CC = gcc -ansi -std=c99
CCFLAGS = -Wall -pedantic -O3
SRCS := $(wildcard *.c)
PRGS := $(patsubst %.c,%,$(SRCS))
all : $(PRGS)
$(PRGS) : $(SRCS)
$(CC) $(CCFLAGS) $< -o $#
This almost works, except it is always grabbing the first dependancy!
..... progA.c -o progA
..... progA.c -o progB
..... progA.c -o progC
So, does anyone have a cleaner method to my 'workaround' or perhaps a solution to my 'near solution'? Is there anyway to match the nth target to the nth dependency when running the command?
Thanks!
The $(PRGS): $(SRCS) change is just incorrect. It lists every source file as the prerequisites of every target.
You want to set the prerequisites of the all target to be every program you want built by default. To do that you want to use all: $(PRGS). Not a recipe of calling make again like you had originally.
(If you did want to keep the manual recursive call to make you could use make --no-print-directory on that call to avoid the message but that is still the wrong approach (and if you were going to do that you'd want to use $(MAKE) to handle arguments to the original make correctly.)
A second problem with your original makefile is the use of *.c in the prerequisite list of the % target. That sets the prerequisites of every target to be every *.c file in the directory. That's not what you want. You want each target to have its own .c file as its prerequisite.
You want:
all: $(PRGS)
%: %.c
$(CC) $(CCFLAGS) $#.c -o $#
That being said make has a built-in rule for exactly that foo.c -> foo compilation so you should just use that. That rule uses the $(CC) and $(CFLAGS) variables. So just set those to what you want and you are done.
So this makefile should do what you want. (Note how I moved some arguments to CFLAGS instead of CC. CC should generally, to my knowledge, by the compiler itself and not any arguments.)
CC = gcc
CFLAGS = -ansi -std=c99 -Wall -pedantic -O3
PRGS := $(patsubst %.c,%,$(wildcard *.c))
all : $(PRGS)
My makefile will not check if there has been any updates and just compiles if it has more than a single source file added in. It works fine with just a single source file.
It seems that it's any source file that's not listed as the first one will always be recompiled and linked.
SOURCES=myclass.cpp mylock.cpp
EXECUTABLE=locktest
LIBRARIES=-pthread
CFLAGS=-Wall
CXX=g++
DIR=host/
EXE=$(EXECUTABLE)
OBJECTS=$(SOURCES:%.cpp=$(DIR)%.o)
$(EXE): $(OBJECTS)
$(CXX) -o $# $(OBJECTS) $(LIBRARIES)
$(DIR)%.o: %.cpp $(DIR)
$(CXX) $(CFLAGS) -c $< -o $#
$(DIR):
#mkdir $(DIR)
clean:
#rm $(OBJECTS) $(EXE)
#rmdir $(DIR)
Output shows problem:
stud#pc:~/Desktop/Locktest$ make
g++ -Wall -c myclass.cpp -o host/myclass.o
g++ -Wall -c mylock.cpp -o host/mylock.o
g++ -o locktest host/myclass.o host/mylock.o -pthread
stud#pc:~/Desktop/Locktest$ make
g++ -Wall -c myclass.cpp -o host/myclass.o
g++ -o locktest host/myclass.o host/mylock.o -pthread
As #lijat points out, when you build an object in $(DIR), the operating system updates the modification time of the directory, so that in this rule:
$(DIR)%.o: %.cpp $(DIR)
...
the prerequisite $(DIR) will always appear to be newer than any target except the last target built.
If your version of Make is recent enough, you can get past this by making $(DIR) an order-only prerequisite:
$(DIR)%.o: %.cpp | $(DIR)
...
Does the filesystem update the modified time on the DIR directory when the compiler writes an .o file there
$(DIR)%.o: %.cpp $(DIR)
ensures that all .o files will be recompiled if anything updates the modified time of that directory.
I need a makefile which get also the name of the file compile
For example:
make foo
and the makefile should compile foo.c to foo.
This is my makefile. How to change it?
all: out
out: out.o
gcc -g -m32 -Wall -o out out.o
out.o: out.c
gcc -m32 -g -Wall -ansi -c -o out.o out.c
.PHONY: clean
#Clean the build directory
clean:
rm -f *.o out
There is no direct way where you can pass arguments to the Makefile but instead you can take advantage of variables to achieve what you want. Check the modifications done to the Makefile below
NAME ?=out #Default binary generated is out if you dont pass any argument
${NAME}: ${NAME}.o
gcc -g -m32 -Wall -o ${NAME} ${NAME}.o
${NAME}.o: ${NAME}.c
gcc -m32 -g -Wall -ansi -c -o ${NAME}.o out.c
.PHONY: clean
#Clean the build directory
clean:
`rm -f *.o ${NAME}`
And you should call the Makefile by typing
$ make NAME=foo
$ make clean NAME=foo
Passing arguments directly to Make is trivially easy.
Your current makefile can be invoked with make foo, and will compile foo.c to produce foo, because Make has implicit rules for handling cases like foo.c => foo; there will be no error even though "foo" is not the target of any rule. (At least, this is the case with GNU Make 3.81, which is what I am using.)
If you want to control the choice of compiler and flags (as in your out rule), there is more than one way to do it. The simplest (though not strictly the best) is to modify a couple of variables in the makefile:
CC = gcc
CFLAGS = -g -m32 -Wall -ansi
Another option is to override the implicit rule with a pattern rule of your own:
%: %.c
gcc -g -m32 -Wall -ansi -o $# $<
If you want it to build foo.o in a separate step, you must split the rule into two rule-- and also put in a rule with no recipe to cancel Make's implicit rule:
%: %.o
gcc -g -m32 -Wall -o $# $^
%.o: %.c
gcc -m32 -g -Wall -ansi -c -o $# $<
%: %.c
Further refinements are possible, once you have mastered the basics.