如何在 Makefile 中正确包含 -Wl,-rpath,$ORIGIN 链接器参数?

2022-01-11 00:00:00 makefile linker c++

我正在使用一些我动态链接的 poco 库在 linux (Ubuntu 16.04) 上准备一个 c++ 应用程序.我的项目文件夹包括:include、bin、lib、src 和 build 文件夹以及相关的 Makefile.到目前为止，我使用了以下 Makefile 从/usr/local/lib 获取库

I'm preparing a c++ app on linux (Ubuntu 16.04) with the use of a few poco libraries which I have dynamically linked. I have project folder that consists of : include, bin, lib , src and build folders and the relevant Makefile. So far I used the following Makefile which got the libraries from /usr/local/lib

CC := g++ 

# Folders
SRCDIR := src
BUILDDIR := build
TARGETDIR := bin

# Targets
EXECUTABLE := C++_APP
TARGET := $(TARGETDIR)/$(EXECUTABLE)

SRCEXT := cpp
SOURCES := $(shell find $(SRCDIR) -type f -name *.$(SRCEXT))
OBJECTS := $(patsubst $(SRCDIR)/%,$(BUILDDIR)/%,$(SOURCES:.$(SRCEXT)=.o))
CFLAGS := -c -Wall
INC := -I include -I /usr/local/include
LIB := -L /usr/local/lib -lPocoFoundation -lPocoNet -lPocoUtil 

$(TARGET): $(OBJECTS)
@echo " Linking..."
@echo " $(CC) $^ -o $(TARGET) $(LIB)"; $(CC) $^ -o $(TARGET) $(LIB)

$(BUILDDIR)/%.o: $(SRCDIR)/%.$(SRCEXT)
@mkdir -p $(BUILDDIR)
@echo " $(CC) $(CFLAGS) $(INC) -c -o $@ $<"; $(CC) $(CFLAGS) $(INC) -c -o $@      $<

clean:
@echo " Cleaning..."; 
@echo " $(RM) -r $(BUILDDIR) $(TARGET)"; $(RM) -r $(BUILDDIR) $(TARGET)

.PHONY: clean

现在我想在运行链接器期间仅在项目 lib 文件夹中搜索库，而不更改 LD_LIBRARY_PATH 或编辑 ld.so.conf.于是搜了一下，发现这可以通过链接器参数-Wl,rpath,$ORIGIN来实现.所以我假设我需要添加以下语句

Now I'd like during running the linker to search for libraries only in project lib folder without changing LD_LIBRARY_PATH or editing ld.so.conf. So I searched and I found that this can be achieved by the linker argument -Wl,rpath,$ORIGIN. So I assume that I need to add the following statement 

LDFLAGS := -Wl,-rpath,$ORIGIN/../lib

并将 LIB 语句更改如下:

and change the the LIB statement as following:

LIB := -L $ORIGIN/../lib -lPocoFoundation -lPocoNet -lPocoUtil

但是它仍然从默认目录 (usr/local/lib) 获取库，因为我在项目 lib 文件夹中没有库的情况下对其进行了测试.我做错了什么?

However it still get the libraries from the default directory (usr/local/lib) , since I tested it with no library on the project lib folder. What have I done wrong?

推荐答案

不，你误会了.您需要将文字字符串 $ORIGIN/../lib 作为参数传递给链接器.$ORIGIN 标记在创建后保存在您的程序中，当运行时链接器开始运行您的程序时，它会将 $ORIGIN 替换为您的程序被调用的当前路径从.即使您将程序复制到其他地方也是如此.因此，如果您将程序作为 /usr/local/bin/myprogram 运行，则运行时链接器会将 $ORIGIN 替换为 /usr/local/bin.如果将其复制到 /opt/mystuff/libexec/myprogram，则运行时链接器会将 $ORIGIN 替换为 /opt/mystuff/libexec.

No, you're misunderstanding. You need to pass the literal string $ORIGIN/../lib as an argument to your linker. The $ORIGIN token is kept inside your program after it's created and when the runtime linker starts to run your program it will replace $ORIGIN with the current path that your program was invoked from. This is true even if you've copied your program somewhere else. So if you run your program as /usr/local/bin/myprogram then the runtime linker will replace $ORIGIN with /usr/local/bin. If you copy it to /opt/mystuff/libexec/myprogram then the runtime linker will replace $ORIGIN with /opt/mystuff/libexec.

为了将文字 $ 传递给 make 配方调用的命令，您必须通过加倍来转义 $:$$.否则，make 会将 $ 视为引入了 make 变量或函数.请记住，make 变量避免括号等是完全合法的，if 它是单个字符(注意，$@、$<等)

In order to pass a literal $ to the command invoked by a make recipe, you have to escape the $ by doubling it: $$. Otherwise, make will see the $ as introducing a make variable or function. Remember, it's perfectly legal for a make variable to avoid the parentheses etc., if it's a single character (note, $@, $<, etc.)

所以当你编写 -Wl,-rpath,$ORIGIN/../lib 时，make 会解释 $ORIGIN 中的 $O作为扩展名为 O 的变量，它是空的，给你 -Wl,-rpath,RIGIN/../lib.

So when you write -Wl,-rpath,$ORIGIN/../lib make will interpret the $O in $ORIGIN as expanding a variable named O, which is empty, giving you -Wl,-rpath,RIGIN/../lib.

您还必须将 $ 从 shell 中转义，否则它将尝试将 $ORIGIN 扩展为您不想要的 shell 变量.

Also you have to escape the $ from the shell, otherwise it will try to expand $ORIGIN as a shell variable which you don't want.

你想做这样的事情:

LDFLAGS = '-Wl,-rpath,$$ORIGIN/../lib' -L/usr/local/lib
LDLIBS = -lPocoFoundation -lPocoNet -lPocoUtil

$(TARGET): $(OBJECTS)
        @echo " Linking..."
        $(CC) $^ -o $@ $(LDFLAGS) $(LDLIBS)

(不知道你为什么用@隐藏命令，然后回显命令...为什么不直接取出@和echo 然后让 make 给你看命令?)

(I don't know why you use @ to hide the command, then echo the command... why not just take out the @ and the echo and let make show you the command?)

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