Even when I do not specify a Makefile, make has built in commands, such as:
$ ls
a.c
$ make a.o
cc -c -o a.o a.c
$
Is there a way to know the exact list of built in commands? For example, I am expecting something like:
%.o: %.c
$(CC) $(CFLAGS) $(AAA) $(BBB) -c -o $# $<
%.o: %.cpp
$(CXX) $(CXXFLAGS) $(AAA) $(BBB) -c -o $# $<
...
It looks like the first flag variable is CFLAGS for C and CXXFLAGS for C++, but I have no idea what the other flag variables are (AAA and BBB above).
Does make store the default rules in some configuration file? Or is it in make's source code?
I am using GNU Make 4.3 on a Fedora Linux machine.
Related
I have Makefile. This runs on FreeBSD with gmake and make. In BSD Make command not output log same with gmake.
$ gmake
compile main.cpp
linking myout
$ make
c++ -O2 -pipe -c main.cpp -o main.o
linking myout
$ cat Makefile
TARGET = myout
default: $(TARGET)
SRCS = main.cpp
OBJS = $(SRCS:%.cpp=%.o)
default: $(BIN)
%.o: %.cpp
#echo compile $<
#$(CXX) -c $< -o $#
$(TARGET): $(OBJS)
#echo linking $#
#$(CXX) $(OBJS) -o $#
clean:
#rm -f $(OBJS) $(TARGET)
According to the FreeBSD make documentation, it doesn't support pattern rules. So your rule here:
%.o: %.cpp
#echo compile $<
#$(CXX) -c $< -o $#
in FreeBSD make is just an explicit rule telling make how to build the literal file %.o from the literal file %.cpp. Since you don't try to build a file named %.o (you're trying to build main.o), this rule is ignored / never used.
It looks like if you want something that will work the same way between both versions of make you'll have to restrict yourself to the POSIX standard suffix rules format, like this:
.SUFFIXES: .cpp .o
.cpp.o:
#echo compile $<
#$(CXX) -c $< -o $#
The default build utilities are different. FreeBSD uses a different implementation of make than GNU/Linux. The respective man pages outline differences.
https://forums.freebsd.org/threads/difference-gmake-gnu-and-freebsd-make.28784/
There are three programs build by this Makefile. They follow the same pattern, but produce different build commands when run. Specifically, I require compilation with c++11 but can only achieve this on one of the build commands. Why is this?
Makefile:
CXX=g++
RM=rm -f
CFLAGS=-std=c++11 -g -Wall $(shell root-config --cflags)
LDFLAGS=-g $(shell root-config --ldflags)
LDLIBS=$(shell root-config --libs)
SOURCES=generic_queue.cpp map_compare.cpp vector_search.cpp
OBJS=$(SOURCES:.cpp=.o)
all: $(SOURCES) generic_queue_test list_of_lists map_compare_test vector_search_test
# Note that $(CFLAGS) is used in the $(CXX) ... command
# each time that a .o file is built.
vector_search_test: $(OBJS) vector_search_test.o
$(CXX) $(LDFLAGS) -o vector_search_test vector_search_test.o $(LDLIBS)
vector_search_test.o: vector_search.cpp vector_search.h
$(CXX) $(CFLAGS) -c vector_search.cpp -o vector_search_test.o
generic_queue_test: $(OBJS) generic_queue_test.o
$(CXX) $(LDFLAGS) -o generic_queue_test generic_queue_test.o $(LDLIBS)
generic_queue_test.o: generic_queue.cpp generic_queue.h fixed_priority_queue.h
$(CXX) $(CFLAGS) -c generic_queue.cpp -o generic_queue_test.o
list_of_lists: $(OBJS) list_of_lists.o
$(CXX) $(LDFLAGS) -o list_of_lists list_of_lists.o $(LDLIBS)
list_of_lists.o: list_of_lists.cpp list_of_lists.h
$(CXX) $(CFLAGS) -c list_of_lists.cpp -o list_of_lists.o
map_compare_test: $(OBJS) map_compare.o
$(CXX) $(LDFLAGS) -o map_compare map_compare.o $(LDLIBS)
map_compare.o: map_compare.cpp map_compare.h
$(CXX) $(CFLAGS) -c map_compare.cpp -o map_compare.o
clean:
$(RM) $(OBJS) generic_queue_test.o list_of_lists.o map_compare.o
dist-clean: clean
$(RM) generic_queue_test list_of_lists map_compare
Output:
g++ -c -o generic_queue.o generic_queue.cpp
g++ -std=c++11 -g -Wall -pthread -m64 -I/usr/include/root -c map_compare.cpp -o map_compare.o
g++ -c -o vector_search.o vector_search.cpp
We see that only the second g++ command fully utilizes CFLAGS variable. Why? Does it have to do with the $(shell ...) portion of the CFLAGS variable?
Edit:
Was able to solve my problem by changing the name of the object file vector_search_test.o to vector_search.o Why did that work?
You have a couple of bugs in your makefile, adding up to this behavior.
First consider OBJS, which contains
generic_queue.o map_compare.o vector_search.o
These files are prerequisites of other targets, but you never actually use generic_queue.o or vector_search.o. Bug #1: you have extra prerequisites by mistake.
These files are prerequisites of other targets, so Make must build them. But how? You have provided rules for three object files:
vector_search_test.o: vector_search.cpp vector_search.h
...
generic_queue_test.o: generic_queue.cpp generic_queue.h fixed_priority_queue.h
...
list_of_lists.o: list_of_lists.cpp list_of_lists.h
...
map_compare.o: map_compare.cpp map_compare.h
...
That last one will do for map_compare.o, but you have given no rules for generic_queue.o or vector_search.o (and there's really no reason you should, since you never use them). But Make knows how to perform certain standard builds, such as foo.cpp => foo.o. If you don't provide a rule, Make will use its implicit rule, which works out to something like this:
generic_queue.o: generic_queue.cpp
$(CXX) $(CPPFLAGS) $(CXXFLAGS) -c generic_queue.cpp -o generic_queue.o
This is very similar to the rules you wrote. In fact, you probably wouldn't have noticed the difference -- and you could have omitted your rules and let Make rely on this one, but Bug #2, you added your flags -std=c++11 -g -Wall whatever to the wrong variable. Make uses CXXFLAGS to hold flags for the C++ compiler; you added yours to CFLAGS, which is for the C compiler.
(I've left out pattern rules and automatic variables since you don't seem to know about them-- I urge you to learn them, they're very useful, but that's for another day.)
i hava a makefile something like this:
outdir = release
allsrc = aaa/a.c bbb/b.c ccc/c.c
allobjs = $(addprefix $(outdir), $(notdir $(allsrc:.c=.o))
...
test: $(allobjs)
$(allobjs): $(allsrc)
gcc -c -o $# $<
make test performs:
gcc -c -o release/a.o aaa/a.c
gcc -c -o release/b.o aaa/a.c
gcc -c -o release/c.o aaa/a.c
(automatic variable $< always takes first prerequisite)
but i want "corresponding one":
gcc -c -o release/a.o aaa/a.c
gcc -c -o release/b.o bbb/b.c
gcc -c -o release/c.o ccc/c.c
what should i change to accomplish desirable result?
i know that this will work for sure:
$(outdir)/a.o: aaa/a.c
gcc -c -o $# $<
$(outdir)/b.o: bbb/b.c
gcc -c -o $# $<
$(outdir)/c.o: ccc/c.c
gcc -c -o $# $<
and wondering how to accomplish the same in one receipe. (because in my real makefile i have ~20 different source files not just 3 like i made here for example)
You don't write your recipe like that. That's not how make works.
That recipe says that every item in $(allobjs) has every item in $(allsrc) as its prerequisite but that's not what you mean.
You mean that every .o file has the matching .c file as its prerequisite which is exactly what the built in %.o: %.c rule already does. You don't even need a makefile to do that.
Edit: Actually, you don't mean that. I had missed that the source files were in different directories. That changes and complicates things.
Chris Dodd wrote up two good solutions for this.
The usual way to do what you are asking would be something like:
outdir = release
allsrc = aaa/a.c bbb/b.c ccc/c.c
allobjs = $(addprefix $(outdir), $(notdir $(allsrc:.c=.o)))
VPATH = $(sort $(dir $(allsrc)))
...
test: $(allobjs)
$(outdir)/%.o: %.c
gcc -c -o $# $<
Of course, this will run into problems if you have a b.c in both aaa and bbb, but since you're trying to put all the object files in the same directory, you have that regardless. It might make more sense to get rid of the $(notdir and keep the same directory structure under $(outdir), in which case you don't need the VPATH
outdir = release
allsrc = aaa/a.c bbb/b.c ccc/c.c
allobjs = $(addprefix $(outdir), $(allsrc:.c=.o))
...
test: $(allobjs)
$(outdir)/%.o: %.c
mkdir -p `dirname $#`
gcc -c -o $# $<
I want to compile my program both to linux and windows using g++ and mingw respectively. The only difference between the compilations is the compiler to use and output file name.
A single make command should produce both output files. What is the best way to achieve this with as little duplications in the makefile as possible?
How about this:
linux-name: CC:=g++
windows-name: CC:=mingw
linux-name windows-name:
$(CC) whatever -o $#
EDIT:
What I wrote above is only the new part of the makefile; I assumed that the rest of the makefile was implied. To be more explicit:
all: linux-name windows-name
linux-name: CC:=g++
windows-name: CC:=mingw
linux-name windows-name: foo.o bar.o baz.o SomethingElse
$(CC) $(CCFLAGS) whatever $^ -o $#
%.o: %.cc
$(CC) $(CFLAGS) -I$(INC_DIR) whatever -c $< -o $#
SomethingElse:
build somehow
I have added another rule to a Makefile to attempt to build a C shared library that uses SWIG to wrap the functions for Java using JNI
The additional rule looks like this (basically lifted from one of the SWIG java examples)
java: $(program_C_SRCS)
$(SWIG) -java $(SWIGOPT) $(INTERFACEPATH)
$(CC) -c $(CFLAGS) $(JAVACFLAGS) $(program_C_SRCS) $(ISRCS) $(CPPFLAGS) $(JAVA_INCLUDE)
$(JAVALDSHARED) $(CFLAGS) $(program_C_OBJS) $(IOBJS) $(JAVA_DLNK) $(LDFLAGS) -o $(JAVA_LIBPREFIX)$(TARGET)$(JAVASO)
javac *.java
problem I have is that my VPATH doesn't seem to work with the *.c files anymore
I noticed that this rule causes all the .c files to compiled as one call to gcc rather than a separate call to gcc for the compilation of each .o file
my previous rules for compilation without any JNI stuff look like this:
.PHONY: all clean
all: $(program_DEBUG_NAME) $(program_RELEASE_NAME)
# original rule to build library in src dir (no longer inc. in all)
$(program_NAME): $(program_C_OBJS)
$(LINK.c) -shared -Wl,-soname,$# $^ -o $#
# new rules to build debug/release libraries and place them in relevant build
# dirs
$(program_DEBUG_NAME): $(DEBUG_OBJS)
$(DEBUG_LINK.c) -shared -Wl,-soname,$# $^ -o $(BUILD_DIR)/debug/$#
$(program_RELEASE_NAME): $(RELEASE_OBJS)
$(RELEASE_LINK.c) -shared -Wl,-soname,$# $^ -o $(BUILD_DIR)/release/$#
# rule to build object files (replaces implicit rule)
$(BUILD_DIR)/debug/%.o: %.c
$(DEBUG_LINK.c) $< -c -o $#
$(BUILD_DIR)/release/%.o: %.c
$(RELEASE_LINK.c) $< -c -o $#
and these work with VPATH no problem
my VPATH statement looks like this:
VPATH = ../../pulse_IO/src ../../../g2/src
Look at your rule:
java: $(program_C_SRCS)
...
$(CC) -c $(CFLAGS) $(JAVACFLAGS) $(program_C_SRCS) ...
...
Suppose program_C_SRCS is foo.c, and the path is somewhere/foo.c. Without VPATH, this rule doesn't run at all because Make can't find foo.c. With VPATH, Make finds it, and knows that the real prereq is somewhere/foo.c. But you have $(program_C_SRCS) in your rule:
java: somewhere/foo.c
...
$(CC) -c $(CFLAGS) $(JAVACFLAGS) foo.c ...
...
This fails because there is no foo.c (locally).
Try this:
java: $(program_C_SRCS)
...
$(CC) -c $(CFLAGS) $(JAVACFLAGS) $^ ...
...
(The use of automatic variables like $^ is the reason your previous rules worked.)