如何在使用gnu-make链接静态库时遵循链接顺序？

Question

I've got the following problem:

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

But

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

runs just fine, nm shows the expected content, tests run, everybody is happy, etc.

I've searched SO and found a plenty of answers (eg Linker order - GCC ), so it's clear that linker works as it really should. So, how should I modify my makefile to follow the order?

Here's the Makefile so far:

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

Tried some weird and obviously wrong things like recursive calling

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

Running this in Debian6 under VirtualBox on Windows7 . System specifications:

$ 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) 

PS its from Zed Shaw's Learn C The Hard Way, exercise 33 . Don't know if I should mark it as a homework :)

Answer 1

You don't show the makefile rule that is building tests/list_tests but it looks as though it's just the built-in rule. With GNU Make, you can print out that rule with -p , which will show you:

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

By adding the library to $(CFLAGS) (via the target-specific variable tests: CFLAGS+=$(TARGET) ), you are placing it before $^ in the resulting command. Instead you should add it to $(LDLIBS) so that it appears after the object files:

tests: LDLIBS+=$(TARGET)

However note that relying on the propagation of target-specific variables like this doesn't work especially well in practice. When you type make tests then the library is used to build tests/list_tests et al. However when you are just interested in one test, you will find that make tests/list_tests fails because the link library is not included in the command. (See this answer for details.)

Answer 2

I'm a noob, progressing through the same book and I got it to build this way:

I changed the line:

tests: CFLAGS+=$(TARGET) #I think this line is useless now

to

tests: CFLAGS+=$(SO_TARGET)