使用 GNU Make 构建具有类似规则的多个可执行文件
问题：
GNU Make 可以吗方便在目录结构中构建具有相似规则的多个可执行文件，从而能够从主目录和各个项目目录进行编译？
答案：
是的，GNU Make 可以处理这项任务采用简洁有效的方法。下面是实现所需功能的两个 makefile：
project.mk:
all :
% : forward_ # build any target by forwarding to the main makefile
$(MAKE) -C .. project_dirs=$(notdir ${CURDIR}) $@
.PHONY : forward_Makefile:
# one directory per project, one executable per directory
project_dirs := $(shell find * -maxdepth 0 -type d )
# executables are named after its directory and go into the same directory
exes := $(foreach dir,${project_dirs},${dir}/${dir})
all : ${exes}
# the rules
.SECONDEXPANSION:
objects = $(patsubst %.cpp,%.o,$(wildcard $(dir )*.cpp))
# link
${exes} : % : $$(call objects,$$*) Makefile
g++ -o $@ $(filter-out Makefile,$^) ${LDFLAGS} ${LDLIBS}
# compile .o and generate dependencies
%.o : %.cpp Makefile
g++ -c -o $@ -Wall -Wextra ${CPPFLAGS} ${CXXFLAGS} -MD -MP -MF ${@:.o=.d} $<
.PHONY: clean
clean :
rm -f $(foreach exe,${exes},$(call objects,${exe})) $(foreach dir,${project_dirs},$(wildcard ${dir}/*.d)) ${exes}
# include auto-generated dependency files
-include $(foreach dir,${project_dirs},$(wildcard ${dir}/*.d))用法：
优点：
以上是GNU Make 能否简化在目录结构中使用类似规则构建大量可执行文件的过程，从而允许从主目录和各个项目文件夹进行编译？的详细内容。更多信息请关注PHP中文网其他相关文章！
