使用SWIG为perl创建C ++接口吗？

Question

I'm trying to grasp SWIG. Right now, I'm failing at building the simple module for perl. Here's what I did:

example.cpp:

#include <iostream>

void hello()
{
    std::cout << "Hello, world!"<< std::endl;
}

example.i:

%module example
%{
#include <iostream>
extern void hello();
%}

extern void hello();

app.pl:

#!/usr/bin/perl

use strict;
use warnings;

use example;

example::hello();

and my compile.sh script:

#!/bin/bash -ex

arch=`perl -e 'use Config; print $Config{archlib};'`    

swig3.0 -c++ -perl5 example.i
g++ -fPIC -c example.cpp
g++ -fPIC -c example_wrap.cxx -I$arch/CORE -L$arch/CORE
g++ -shared example_wrap.o -o example.so

Compiles fine - no errors are shown. However, when I run app.pl , I get following error:

/usr/bin/perl: symbol lookup error: ./example.so: undefined symbol: _Z5hellov

I tried to google a bit, but to no avail. So far, everything has failed. What am I doing wrong?

I'm using Linux Ubuntu x86_64 with 3.13 kernel, and perl 5.18.2

Answer 1

Behold the power of the small break.

The reason is dead simple: I didn't add example.o to the compile chain - so there indeed was no definition of hello() function!

To solve it, I changed compile.sh last line to this: g++ -shared example.o example_wrap.o -o example.so

I feel stupid.