I would like to compile a binary that doesnot depend on LD_LIBRARY_PATH as this binary will be setuid to owner and used by other users and since setuid doesnot support LD_LIBRARY_PATH making it independent of LD_LIBRARY_PATH would be great.
But I am not able to specify the path of the shared libraries to the linker at compile time. I am using gcc compiler 4.1.2 and on Linux OS Red Hat 5.8. I am using the following compile command where /aaa/bbb/lib is the path of the shared library that is used by the binary waitdb.ORACLE called within the binary simple:
gcc simple.c -Wl,-rpath=/aaa/bbb/lib -o simple
simple.c has nothing but an execvp call to another binary 'waitdb.ORACLE' which uses libuidata.so that is located in /aaa/bbb/lib directory
Contents of simple.c are as follows:
#include <stdio.h>
int main(int argc, char* argv[])
{
char* args[3];
args[0]="/aaa/bbb/bin/waitdb.ORACLE";
args[1]=NULL;
printf ("before executing execvp\n");
execvp (args[0],args);
printf ("after executing execvp\n");
}
But when I run the executable 'simple' I get the following error
[acdev2@hostname tmp]$ ./simple
before executing execvp
/aaa/bbb/bin/waitdb.ORACLE: error while loading shared libraries: libuidata.so: cannot open shared object file: No such file or directory
On the other hand if I run at the commandline, the export LD_LIBRARY_PATH command before calling the binary it works FINE.
[acdev2@hostname tmp]$ export LD_LIBRARY_PATH=$LD_LIBRARY_PATH:/aaa/bbb/lib
[acdev2@hostname1 tmp]$ ./simple
before executing execvp
+++INFO+++ 20140103_13:54:32 @(#)waitdb[6.2/$Revision: 5538 $]: RDBMS DNYASC01 is accepting connections
Please advise how I can specify the /aaa/bbb/lib path to the linker at compile time of binary 'simple' so that when 'simple' binary tries to execvp another c binary 'waitdb.ORACLE' binary within it the linker knows where to find the libuidata.so shared library file that waitdb.ORACLE uses (without using the LD_LIBRARY_PATH variable) ?
thanks