如何使用 GNU Make 从不同目录高效构建具有相似规则的多个可执行文件？-C++-PHP中文网

如何使用 GNU Make 从不同目录高效构建具有相似规则的多个可执行文件？

Linda Hamilton

发布： 2024-10-29 08:42:03

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How can GNU Make be used to efficiently build multiple executables with similar rules from different directories?

使用 GNU Make 构建具有类似规则的可执行文件

作为 Scons 的替代方案，GNU Make 可以成为您任务的有效解决方案。 GNU Make 允许您通过其依赖解析系统定义灵活的构建过程。

GNU Make 方法

我们提供了两个 makefile，可以从顶部进行构建和清理-级别目录（all_lessons）和单个项目目录：

project.mk

all :
% : forward_ # build any target by forwarding to the main makefile
    $(MAKE) -C .. project_dirs=$(notdir ${CURDIR}) $@
.PHONY : forward_

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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))

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