在C ++库中创建对象（如std :: cout）

Question

I recently read somewhere that std::cout is an instance of the std::ostream class. I wish to implement a similar kind of thing. I make a class Animal and want to provide an instance of Animal class dog in the library itself like std::cout . I am not sure how to do it but here's a code snippet which hopefully will give you an idea of what I'm trying to achieve.

// lib.h

#ifndef LIB_H
#define LIB_H

#include <string>

class Animal {
public:
    Animal();
    std::string name;
};

Animal dog;
dog.name = "dog";
extern Animal dog;

#endif

---

// lib.cpp

#include "lib.h"

Animal::Animal() {}

Animal dog;
dog.name = "dog";

---

// main.cpp

#include "lib.h"
#include <iostream>

int main() {
    Animal my_dog = dog;
    std::cout << my_dog.name << std::endl;
    return 0;
}

This is the error I get when I try this code:

lib.cpp:6:1: error: ‘dog’ does not name a type
 dog.name = "dog";

This kind of code may seem silly, but I still am looking for ways to implement this approach and not its alternatives.

EDIT : After several answers which are suggesting using inheritance or including the parameters in constructor. I will explain the actual problem that I am dealing with.

I am working on a motion planning library. The library has an object called RoadmapPlanner . It is something similar to what is shown below

class RoadmapPlanner {
private:
    Graph g;
public:
    std::function<Graph(Workspace)> build_graph;
    std::function<Path(Graph,Point,Point) find_path;
    // and several others
}

Now I want to provide some pre-built motion planners to the user. For example: I want to provide the voronoi_planner , which I want to implement by using the following syntax:

RoadmapPlanner voronoi_planner;
voronoi_planner.build_graph = voronoi_graph_builder; // this describes what algorithms
voronoi_planner.graph_search = a_star_search;        // will be used internally
voronoi_planner.load_map("map_001.png");
voronoi_planner.run();

and so on... voronoi_graph_builder , a_star_search etc are functions that I have already written. This thing works perfectly when I do this in the main function. I want to provide several pre-built planners and also allow the user to build planner themselves by using the above syntax. The pre-built planners must be a part of the library so the user can use them by following the below syntax:

RoadmapPlanner my_planner = voronoi_planner;
my_planner.search = rrt_search;

I hope this explains what I intend to do.

Answer 1

You cannot do the assignment of dog.name outside any function, you need to construct your Animal with its name directly:

// lib.hpp

class Animal {
public:
    Animal(std::string);
    std::string name;
};

extern Animal dog;

// lib.cpp

Animal::Animal(std::string name) : name(name) {}

Animal dog("dog");

If your type meets the requirements for aggregate initialization , you can omit the constructor:

// lib.hpp

class Animal {
public:
    std::string name;
};

extern Animal dog;

// lib.cpp

Animal dog = {"dog"};

Update after your edit:

In your edit, you define the variable dog in both lib.cpp and lib.hpp , this cannot work per the one definition rule : dog will be define in each of the .cpp file which include lib.hpp (and thus will be define multiple times).

If you want to declare dog in your lib.hpp , you should use an anonymous namespace and not define it in lib.cpp .

Answer 2

C++ does not allow statements to be placed in the global scope. The simplest way to solve this is to make Animal take its name in in its constructor so you can init your dog like so:

Animal dog("dog");

Also consider placing your static instance inside the Animal class itself, this way you get the added benefit of proper scope names.

// .h

class Animal {
public:
    Animal();
    std::string name;
    static Animal dog;
};

// .cpp
Animal Animal::dog("dog");

Answer 3

int i;

struct init_i
{
    init_i()::init_i() { i = 5; }
};

init_i _init;

i will be set to 5 before main()

Answer 4

What your looking for is class inheritance. In your case it would look like this:

// lib.h

class Animal {
public:
    Animal();
    std::string name;
};

class Dog : public Animal {
public:
    Dog();
};

// lib.cpp

Animal::Animal() {}
Dog::Dog() { name = "dog"; }

// main.cpp

#include "lib.h"

int main() {
    Animal my_dog = Dog();
    std::cout << my_dog.name << std::endl;
    return 0;
}

Answer 5

I suggest you read about Object Oriented Programming Paradigms (Encapsulation, Inheritance Polymorphism)

Encapsulation means you will keep the attributes of your objects hidden from outside the class and use public method to assign or retrieve values from the object.

Class Animal // parent class
{
    protected:
        std::string name;
    public:
        void setName(std::string n) { name = n; }
        std::string getName() { return name; }
}

Inheritance means allowing an object (child object) to get the same attributes form its parents. In addition you can have additional attributes and methods unique to the child object.

Class Dog: Animal   // child class inheriting parent class
{
    std::string sound;
    public:
        // now you can call getName and setName methods for Dog objects.
        void setSound(std::string s) { sound = s; }
        std::string getSound() { return sound; }
}

Usage:

Dog Askal;

Askal.setName("Bantay");
Askal.setSound("Woof");

Polymorphism allows the inherited method of a parent to change its behavior (code) according to its derived class. I leave this part for you to research since its not even required (yet) in your problem.

Reply to your EDIT: Why not use "enum" to list down different planners and include it in your library. Then use a switch statement to search what particular planner the user is looking for then execute the corresponding function (or algorithm)