Hi,

I am trying to compile a simple shared library on my linux box. I m running RHEL 5.

This is the code for the lirary file
//-----------------libhello.c---------
#include <stdio.h>

void sayhello(){
printf("library saying hello!! \n");
}

this is the header file
//--------------libhello.h------

//declaring the functions!!

void sayhello();

I have compiled the header source file as
gcc -fPIC -c -o libhello.o libhello.c

and then, to make it a shared library,
gcc -shared -Wl,-soname,libhello.so.1 -o libhello.so.1.0.0 libhello.o

my PWD: /home/harsha/lib

then I used ldconfig as
/sbin/ldconfig -v -n . (output shown below)
.:
libhello.so.1 -> libhello.so.1.0.0

and I have set the LD_LIBRARY_PATH env var as
export LD_LIBRARY_PATH=/usr/harsha/lib

now I wrote a simple program to used the above created shared library
//--------------hello.c-----------------
#include "libhello.h"

int main(void)
{
sayhello();
}

and compiled as
gcc -c -o hello.o hello.c

now when I try to link the above object file and the shared library I am facing this error..
gcc -o dynamic_out hello.o -L. -lhello
/usr/bin/ld: cannot find -lhello
collect2: ld returned 1 exit status

the soname link has been created by the ldconfig and somehow this error is showing up..

Welcome to Linux Questions.

Very simple :
'ln -s libhello.so.1 libhello.so'

Reason : If you have e.g.
libhello.so.1.0.0, libhello.so.2.0.0, libhello.so.3.0.0
it will be your decision, which one to compile against.

So the linker ( ld ) will use "lib.so" only.
....

Thanks pal..

Creating the symbolic link worked.. I was little confused before understanding how ld determines which library version to use when multiple versions of the same library exits.. but now, your post clarifies a lot!

Kudos.