CODEWITHSUNDEEP
c++stlreturnexportpimpl-idiom
Jonathan Mee
Jonathan Mee

Reputation: 38919

PImpl Doesn't Save Me from having to Export STL

I pursued the PImpl design to avoid having to export STL from my dynamic library.

Old:

//In header file
class Foo{
public:
    const map<char, char>& getMap() {return _map;}
private:
    map<char, char> _map;
};

New:

//In header file
class Foo{
public:
    Foo();
    ~Foo();
   const map<char, char>& getMap();
private:
    struct Impl;
    Impl* _pimpl;
};

//In implementation file
struct Foo::Impl{
    map<char, char> _map;
}
Foo::Foo(): _pimpl(new Impl){}
Foo::~Foo(){delete _pimpl;}
const map<char, char>& Foo::getMap(){return _pimpl->_map;}

However the clear issue is that I still have to export the map as part of my library. I don't want to stop returning STL, but I don't see a way around it. Is there another paradigm which will still let me return STL but not have to export it?

Upvotes: 1

Views: 154

Answers (1)

juanchopanza
juanchopanza

Reputation: 227418

A solution is to not use std::map in the interface of your class, but rather, implement the requires subset of methods. For example, assuming you want read-write element access via operator[],

// Foo.h
class Foo{
public:
    Foo();
    ~Foo();
   const char& operator[](char key) const;
   char& operator[](char key);
private:
    struct Impl;
    Impl* _pimpl;
};

then

// Foo.cpp
#include <map>
struct Foo::Impl
{
    std::map<char, char> _map;
};

Foo::Foo() : _pimpl(new Impl){}

Foo::~Foo(){delete _pimpl;} // don't forget to take care of copy and assignment

const char& Foo::operator[](char key) const {return _pimpl->_map[key];}

char& Foo::operator[](char key) {return _pimpl->_map[key];}

Upvotes: 1

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