这个 Makefile 构建了三个程序。它们遵循相同的模式,但在运行时会产生不同的构建命令。具体来说,我需要编译,c++11
但只能在其中一个构建命令上实现。为什么是这样?
生成文件:
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
$(RM) generic_queue_test list_of_lists map_compare
输出:
g++ -c -o generic_queue.o generic_queue.cpp
g++ -std=c++11 -g -Wall -pthread -m64 -I/usr/include/root -c map_compare.cpp -o map_compare.o
g++ -c -o vector_search.o vector_search.cpp
我们看到只有第二个g++
命令充分利用了CFLAGS
变量。为什么?它与变量的$(shell ...)
部分有关吗?CFLAGS
编辑:能够通过将目标文件的名称更改vector_search_test.o
为vector_search.o
为什么可以解决我的问题?