我正在尝试学习SWIG,我遇到一些问题,让SWIG与
Linux机器上的perl一起工作.我有文件Dog.h,Crow.h,Animal.i和libmylib.so.所有这些文件都在同一目录中. libmylib.so使用Dog.cpp和Crow.cpp编译,分别引用Dog.h和Crow.h.我的Animal.i文件如下：
%module Animal
%{
/* Includes the header in the wrapper code */
#include "Dog.h"
#include "Crow.h"
%}
/*Parse the header file to generate wrappers */
%include "Dog.h"
%include "Crow.h"
以下是我为执行构建perl模块而执行的命令：
swig -perl -c++ Animal.i g++ -shared -fPIC Animal_wrap.cxx -L. -lmylib -I/usr/lib64/perl5/CORE -o _Animal.so LD_LIBRARY_PATH=. perl
当我输入“use Animal;”时,我收到以下错误：“无法在@INC中找到模块Animal的可加载对象”.我对perl相当新,所以我不知道如何解决这个问题,虽然从搜索中我觉得问题可能是perl无法引用我的libmylib.so文件.任何帮助将不胜感激.谢谢！
解决方法
以下似乎适用于Ubuntu 16.04：
文件：
Animal.i：
%module Animal
%{
#include "Dog.h"
#include "Crow.h"
%}
%include "Dog.h"
%include "Crow.h"
Crow.h
class Crow {
public:
Crow() {
ncrows++;
}
virtual ~Crow() {
ncrows--;
}
static int ncrows;
};
Dog.h：
class Dog {
public:
Dog() {
ndogs++;
}
virtual ~Dog() {
ndogs--;
}
static int ndogs;
};
Crow.cpp：
#include "Crow.h" int Crow::ncrows = 0;
Dog.cpp：
#include "Dog.h" int Dog::ndogs = 0;
test.pl：
use strict; use warnings; use Animal; print "Creating a Crow:\n"; my $c = Animal::Crow->new(); print " Created crow $c\n"; $c->DESTROY(); print "Creating a Dog:\n"; my $d = Animal::Dog->new(); print " Created dog $d\n"; $d->DESTROY();
汇编：
swig -perl -c++ Animal.i g++ -fPIC -c Crow.cpp g++ -fPIC -c Dog.cpp g++ -shared Crow.o Dog.o -o libmylib.so g++ -fPIC -c Animal_wrap.cxx -I/usr/lib/x86_64-linux-gnu/perl/5.22/CORE g++ -shared -L. Animal_wrap.o -lmylib -o Animal.so
运行测试脚本：
$LD_LIBRARY_PATH=. perl test.pl
Creating a Crow:
Created crow Animal::Crow=HASH(0x10c2eb0)
Creating a Dog:
Created dog Animal::Dog=HASH(0x10c2f88)