g++ 未定义的引用，尽管 *.so 文件中存在符号 [英] g++ undefined reference although symbol is present in *.so file
问题描述
我发现了许多类似的问题(例如 这个，那个 或 this)，但它们都没有帮助我解决我的问题.我有一个 *.so 文件(来自 gnss-sdr 的核心)，如下所示:
I found a number of similar questions (e.g. this, that or this), but none of them helped me solve my problem. I have a *.so file (from the core of gnss-sdr) that, as indicated by:
$nm libgnss_system_parameters_dyn.so | c++filt |grep Gps_Eph
包含符号Gps_Ephemeris::Gps_Ephemeris()，应该是构造函数.
contains the symbol Gps_Ephemeris::Gps_Ephemeris(), which is supposed to be a constructor.
我写了一些最少的代码:
I've written some minimal code:
#include <iostream>
#include <core/system_parameters/gps_ephemeris.h>
int main(int argc,const char* argv[])
{
Gps_Ephemeris ge;
return 0;
}
我用来编译的:
g++ main.cpp -std=c++0x -I some_include_path -L some_lib_path -l gnss_system_parameters_dyn`
然后链接器抱怨:
/tmp/ccHCvldG.o: In function `main':
main.cpp:(.text+0x33): undefined reference to `Gps_Ephemeris::Gps_Ephemeris()'
collect2: error: ld returned 1 exit status
我也试过cmake，但是它生成的那一行是类似的(它只是在链接之前添加了-rdynamic)，它仍然产生了完全相同的链接器错误.
I also tried cmake, but the line it generated was similar to that (it just added -rdynamic before linking), and it still generated the exact same linker error.
请注意，库和我的最小代码都是使用相同的编译器 (g++-5) 编译的，具有完全相同的标志和相同的 c++0x 标准.
Note that both the library and my minimal code are being compiled with the same compiler (g++-5), with the exact same flags and the same c++0x standard.
Maxim Egorushkin 的回答如下:
Addressing the answer by Maxim Egorushkin, the line:
nm --demangle --defined-only --extern-only libgnss_system_parameters.so |grep Gps_Eph
不输出任何东西.但是，符号是在静态库中定义的(即 *.a 库):
doesn't output anything. However, the symbol is defined in the static library (i.e. the *.a library):
00000000000006b0 T Gps_Ephemeris::Gps_Ephemeris()
00000000000006b0 T Gps_Ephemeris::Gps_Ephemeris()
知道两者都是由cmake生成的，方式如下:
Knowing that both are generated by cmake, in the following way:
add_library(lib_name SHARED ${sources_etc}) #for the *.so
add_library(lib_name_2 ${sources_etc}) #for the *.a
这些库中包含/定义的符号应该没有区别，对吧?我没有注意到 cmake 关于 add_library 的文档中的任何内容.我错过了什么明显的东西吗?
there should be no difference in symbols contained/defined in those libraries, right? I didn't notice anything in cmake's documentation on add_library. Am I missing something obvious?
推荐答案
检查 .so 是否导出符号的迂腐正确方法是 nm --demangle --dynamic --仅定义 --extern-only <lib.so>|grep <符号>.
The pedantically correct way to check that a .so exports a symbol is nm --demangle --dynamic --defined-only --extern-only <lib.so> | grep <symbol>.
如果没有 --defined-only，您的命令也会显示 undefined 符号.
Without --defined-only your command also shows undefined symbols.
如果没有 --extern-only，它还会显示带有 internal links 的符号，这些符号无法链接.
Without --extern-only it also shows symbols with internal linkage which are unavailable for linking.
您似乎需要链接另一个库，因为链接 libgnss_system_parameters_dyn.so 无法解析 Gps_Ephemeris::Gps_Ephermeris().一个好的开始方式是该库的文档和示例.
It looks like you need to link another library because Gps_Ephemeris::Gps_Ephermeris() is not resolved by linking libgnss_system_parameters_dyn.so. A good way to start is that library's documentation and examples.
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