如何在使用gnu-make链接静态库时遵循链接顺序?

我有以下问题:

cc -g -O2 -Wall -Wextra -Isrc -rdynamic -DNDEBUG  build/liblcthw.a    tests/list_tests.c   -o tests/list_tests
/tmp/ccpvGjZp.o: In function `test_create':
~/lcthw/tests/list_tests.c:12: undefined reference to `List_create'
collect2: ld returned 1 exit status
make: *** [tests/list_tests] Error 1

cc -g -O2 -Wall -Wextra -Isrc -rdynamic -DNDEBUG tests/list_tests.c  build/liblcthw.a -o tests/list_tests

运行得很好,nm显示预期的内容,测试运行,每个人都很高兴,等等.

我搜索了SO并找到了大量的答案(例如Linker order – GCC),所以很明显链接器的工作原理应该如此.那么,我应该如何修改我的makefile以遵循命令呢?

到目前为止,这是Makefile:

CFLAGS=-g -O2 -Wall -Wextra -Isrc -rdynamic -DNDEBUG $(OPTFLAGS)
LIBS=$(OPTLIBS)
PREFIX?=/usr/local
BUILD=build

SOURCES=$(wildcard src/**/*.c src/*.c)
OBJECTS=$(patsubst %.c,%.o,$(SOURCES))

TEST_SRC=$(wildcard tests/*_tests.c)
TESTS=$(patsubst %.c,%,$(TEST_SRC))

TARGET=$(BUILD)/liblcthw.a
TARGET_LINK=lcthw
SO_TARGET=$(patsubst %.a,%.so,$(TARGET))

#The Target Build
all: $(TARGET) $(SO_TARGET) tests

dev: CFLAGS=-g -Wall -Isrc -Wall -Wextra $(OPTFLAGS)
dev: all

$(TARGET): CFLAGS += -fPIC
$(TARGET): build $(OBJECTS)
    ar rcs $@ $(OBJECTS)
    ranlib $@

$(SO_TARGET): $(TARGET) $(OBJECTS)
    $(CC) -shared -o $@ $(OBJECTS)

build:
    @mkdir -p $(BUILD)
    @mkdir -p bin

#The Unit Tests
.PHONY: tests
tests: CFLAGS+=$(TARGET)     #I think this line is useless now
tests: $(TESTS)
    sh ./tests/runtests.sh

#some other irrelevant targets

尝试了一些奇怪的,明显错误的事情,如递归调用

$(TESTS):
    $(MAKE) $(TESTS) $(TARGET)

在Windows7上的VirtualBox下的Debian6中运行它.系统规格:

$uname -a
Linux VMDebian 2.6.32-5-686 #1 SMP Mon Mar 26 05:20:33 UTC 2012 i686 GNU/Linux
$gcc -v
Using built-in specs.
Target: i486-linux-gnu
Configured with: ../src/configure -v --with-pkgversion='Debian 4.4.5-8' --with-bugurl=file:///usr/share/doc/gcc-4.4/README.Bugs --enable-languages=c,c++,fortran,objc,obj-c++ --prefix=/usr --program-suffix=-4.4 --enable-shared --enable-multiarch --enable-linker-build-id --with-system-zlib --libexecdir=/usr/lib --without-included-gettext --enable-threads=posix --with-gxx-include-dir=/usr/include/c++/4.4 --libdir=/usr/lib --enable-nls --enable-clocale=gnu --enable-libstdcxx-debug --enable-objc-gc --enable-targets=all --with-arch-32=i586 --with-tune=generic --enable-checking=release --build=i486-linux-gnu --host=i486-linux-gnu --target=i486-linux-gnu
Thread model: posix
gcc version 4.4.5 (Debian 4.4.5-8) 

附:它来自Zed Shaw的Learn C The Hard Way,exercise 33.不知道我是否应该把它标记为作业:)

最佳答案
您没有显示构建tests / list_tests的makefile规则,但它看起来好像只是内置规则.使用GNU Make,您可以使用-p打印出该规则,它将显示:

# default
LINK.c = $(CC) $(CFLAGS) $(CPPFLAGS) $(LDFLAGS) $(TARGET_ARCH)
[...]
.c:
#  recipe to execute (built-in):
    $(LINK.c) $^ $(LOADLIBES) $(LDLIBS) -o $@

通过将库添加到$(CFLAGS)(通过特定于目标的变量测试:CFLAGS = $(TARGET)),您将它放在结果命令中的$^之前.相反,你应该将它添加到$(LDLIBS),以便它出现在目标文件之后:

tests: LDLIBS+=$(TARGET)

但请注意,依赖于像这样的目标特定变量的传播在实践中并不是特别有效.当您键入make tests时,该库将用于构建tests / list_tests等.但是,当您只对一个测试感兴趣时,您会发现make tests / list_tests失败,因为该命令中不包含链接库. (详见this answer)

转载注明原文:如何在使用gnu-make链接静态库时遵循链接顺序? - 代码日志