Export a class with std::string

Question

I know this subject has been covered and re-talked, but I still get stuck every time I need to do something like that, and the internet is full of different answers. so I decided to simply ask how to deal with such situation once and for all.

Lets say I have the following class:

class PETS_EXPORT_API dog
{
public:
  dog(std::string name):_name(name){}
  ~dog(){}

private:
  std::string _name;
};

Obviously this code would generate a warning because I'm trying to export std::string. How do I solve such issue ?

thanks!

Answer 1

Alternative to Joe McGrath's answer:

If you really want your clients to have access to Dog public & protected interface, and does not make sense to have an abstract interface,

• You could use the pImpl idiom, to hide the private interface.
• Additionally you could export the string in the form of chars

Hide your original dog:

class Dog_Impl // your original class
{

public:
  Dog_Impl(std::string name):_name(name){}
  ~Dog_Impl(){}

  string::get_name();
private:
  std::string _name;

};

Put this into your API:

class Dog_Impl; // fwd declaration

class PETS_EXPORT_API Dog {
  public:
    Dog(const char *name);
    ~Dog();
    const char *get_name();

  private:
    Dog_Impl *pImpl;
};

The implementation should simply pass all public & protected interface to the pImpl:

Dog::Dog(const char *name) 
{
  pImpl = new Dog_Impl(name);
}

Dog::~Dog()
{ 
  delete pImpl;
}

const char *Dog::get_name()
{
  return pImpl->get_name().c_str(); 
}