GNU Make 能否簡化在目錄結構中使用類似規則建構大量可執行檔的過程，從而允許從主目錄和各個專案資料夾進行編譯？-C++-PHP中文網

GNU Make 能否簡化在目錄結構中使用類似規則建構大量可執行檔的過程，從而允許從主目錄和各個專案資料夾進行編譯？

Linda Hamilton

發布： 2024-10-27 16:07:02

原創

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Can GNU Make streamline building numerous executables with similar rules across a directory structure, allowing compilation from both the main directory and individual project folders?

使用GNU Make 構建具有類似規則的多個可執行文件

問題：

GNU Make 可以嗎方便在目錄結構中建立具有相似規則的多個可執行文件，從而能夠從主目錄和各個專案目錄進行編譯？

答案：

是的，GNU Make 可以處理這項任務採用簡潔有效的方法。以下是兩個實現所需功能的makefile：

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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Makefile: