c++ - linux下C++中无需安装的共享对象和包含

Question

我正在编写一个程序，其中有两个我需要使用的库:v8 和 v8-juice。不幸的是，v8-juice 不能编译为静态库，因为它与模板有关。它还有一些其他怪癖，需要将 v8 编译为共享对象。

因此，当我编译我的程序时，我最终得到了运行可执行文件所需的两个共享对象。我的问题是，有没有一种方法可以包含这些共享对象而无需在 linux 下安装它们？抱歉，如果这是一个新问题，我是 C++ 的新手。

Answer 1

共享库可以与您的可执行文件位于同一文件夹中。 man ld.so:

   $ORIGIN and rpath

   ld.so  understands the string $ORIGIN (or equivalently ${ORIGIN}) in an
   rpath specification (DT_RPATH or DT_RUNPATH) to mean the directory con-
   taining  the  application  executable.  Thus, an application located in
   somedir/app could be compiled with gcc  -Wl,-rpath,'$ORIGIN/../lib'  so
   that  it  finds  an  associated shared library in somedir/lib no matter
   where somedir is located in the directory hierarchy.  This  facilitates
   the  creation  of  "turn-key"  applications  that  do  not  need  to be
   installed into special directories, but can instead  be  unpacked  into
   any directory and still find their own shared libraries.