Question

我有一个C ++项目的以下目录结构：

.
├── bin
├── build
├── include
│   ├── dir1
│   │   ├── file1.hpp
│   │   └── file2.hpp
│   ├── dir2
│   │   ├── file3.hpp
│   │   └── file4.hpp
│   └── third_party
│       └── catch.hpp
├── Makefile
├── src
│   ├── dir1
│   │   ├── file1.cpp
│   │   └── file2.cpp
│   └── dir2
│       ├── file3.cpp
│       └── file4.cpp
└── test
    ├── dir1
    │   ├── file1.test.cpp
    │   └── file2.test.cpp
    └── dir2
        ├── file3.test.cpp
        └── file4.test.cpp

如何编写Makefile来编译src和test目录中的代码，并获取build目录中的目标文件和bin目录中的二进制文件，在其中维护相同的目录结构，，而不必明确命名每个文件及其依赖？在dir*和src中的每个test中使用多个Makefile会更好吗？

我的Makefile目前看起来像这样:(这可能是荒谬的，对不起！）

binaries_dir = bin
build_dir = build
sources_dir = src
include_dir = include

compile_flags = -std=c++14 -Wall

binaries := $(wildcard *.out)
objects := $(wildcard *.o)
sources := $(wildcard *.cpp)
headers := $(wildcard *.hpp)

objects: $(sources)
    g++ $(compile_flags) -c $(sources_dir)/$(sources) -I $(include_dir)

binaries: $(objects)
    for object in $(objects); do
        g++ $(compile_flags) -o $(binaries_dir)/ $(build_dir)/$object
    done

Answer 1

我认为这不是万无一失的，但我认为它几乎可以满足您的需求。你应该研究一下Makefiles的构建，因为它的结果并不完全符合你的需求，你将如何修复它？

CXX := g++
RM  := rm -f
MD  := mkdir -p

# don't change, this is for dependencies
CXXFLAGS += -MMD -MP

# add compiler flags here
CXXFLAGS += -std=c++14 -pedantic-errors
CXXFLAGS += -Wall -Wextra
CXXFLAGS += -g3 -O0

# add external includes here
CPPFLAGS += -Iinclude

# add library flags here
LDFLAGS +=

DIRS := $(patsubst  src/%, %, $(wildcard  src/*))
DIRS += $(patsubst test/%, %, $(wildcard test/*))
PROG_SOURCES := $(wildcard  src/*/*.cpp)
TEST_SOURCES := $(wildcard test/*/*.cpp)
OBJECTS := $(patsubst  src/%.cpp, build/%.o, $(PROG_SOURCES))
OBJECTS += $(patsubst test/%.cpp, build/%.o, $(TEST_SOURCES))
EXECUTABLES := $(patsubst  src/%.cpp, bin/%, $(PROG_SOURCES))
EXECUTABLES += $(patsubst test/%.cpp, bin/%, $(TEST_SOURCES))
DEPENDENCIES := $(patsubst   src/%.cpp, build/%.d, $(PROG_SOURCES)) 
DEPENDENCIES += $(patsubst  test/%.cpp, build/%.d, $(TEST_SOURCES)) 

all: dirs $(EXECUTABLES)

build/%.o: src/%.cpp
    $(CXX) -c $(CXXFLAGS) $(CPPFLAGS) -o $@ $<

build/%.o: test/%.cpp
    $(CXX) -c $(CXXFLAGS) $(CPPFLAGS) -o $@ $<

bin/%: build/%.o
    $(CXX) $(CXXFLAGS) -o $@ $< $(LDFLAGS)

-include $(DEPENDENCIES)

clean:
    @echo Removing build files
    @$(RM) $(EXECUTABLES) $(OBJECTS) $(DEPENDENCIES)

dirs:
    @$(MD) $(patsubst %, build/%, $(DIRS)) $(patsubst %, bin/%, $(DIRS))

.PHONY: show dirs

Answer 2

您可能会从中找到一些灵感：https://github.com/jschmerge/DasBuild

包含多个目录但相同目录结构的代码的项目Makefile

2 个答案: