c + +使用一个类中的一类（一类），并调用其功能

Question

我有以下问题：我想从好老C项目重做C++，使一切class(y) :)并保持它从一开始可扩展性。

它是细胞的网格上的一个模拟（A群被部分），所以决定以下结构：

class Simulation has an instance of 
class Grid  has an instance of 
class Swarm  has an instance of 
class Cell 

我在单独的头文件中定义的类。那么当然，我需要能够在grid，swarm和cell中调用函数。我想要做的直截了当：

Simulation mysim; 
mysim.get_grid(0).any_function_here(); 

与电网作为返回参数

Grid Sim::get_grid(int grid_no) 
{ 
    std::cout << "sim.get_grid(" << grid_no << ") called." << std::endl; 
    if (grid_no <= amount_of_grids) 
     return this->test;//##//this->gridlist[grid_no]; 
    else 
     std::cout << "you have not created this grid number yet" << std::endl; 

    Grid dummy; 
    return dummy; 
} 

它调用的函数，并且只要在电网未进行任何更改的作品。这些似乎在太空中消失了。可能是一个指针错误，但我不能找到一个错误，因为完全相同的代码工作的Simulation类...

更多来源：

int Grid::create_swarm(std::string name) 
{ 
    Swarm new_swarm; 
    new_swarm.set_name("Protoswarm"); 
    swarmlist.push_back(new_swarm); 
    this->amount_of_swarms ++; 
    std::cout << "amount_of_swarms = " << amount_of_swarms << std::endl; 
    return 0; 
} 

Swarm Grid::get_swarm(int swarm_no) 
{ 
    std::cout << "grid.get_swarm(" << swarm_no << ") called." << std::endl; 
    if (swarm_no <= amount_of_swarms) 
     return swarmlist[swarm_no]; 
    else 
     std::cout << "oh oh - you have not this swarm in here..." << std::endl; 

    Swarm dummy; 
    return dummy; 
} 

我可以随时调用create_swarm功能我想要，但是群体永远不会出现，并且柜台不会在该网格中提升，只要功能在那里即可。我错过了什么吗？它真的只是一个指针错误？为什么这段代码工作，如果我这样称呼它：

Grid newgrid; 
newgrid.create_swarm(); 

一个快速ç& p'ed MWE

#include <iostream> 
#include <string> 
#include <vector> 

class Sim 
{ 
    public: 
     Sim(); 
     virtual ~Sim(); 

     Grid get_grid(int grid_no); 

    protected: 

    private: 
     std::vector<Grid> gridlist; 
     int amount_of_grids = -1; 
}; 


class Grid 
{ 
    public: 
     Grid(); 
     virtual ~Grid(); 

     int set_size(int x, int y); 
     int create_swarm(std::string name); 
     Swarm get_swarm(int swarm_no); 
     void print_swarms(); 

    protected: 

    private: 
     std::vector<Swarm> swarmlist; 
     int amount_of_swarms = -1; 
     /*static const*/ int size_x; 
     /*static const*/ int size_y; 
     std::vector<std::vector<Field>> fields; 
     std::string gridname; 
}; 

Grid Sim::get_grid(int grid_no) 
{ 
    std::cout << "sim.get_grid(" << grid_no << ") called." << std::endl; 
    if (grid_no <= amount_of_grids) 
     return this->gridlist[grid_no]; 
    else 
     std::cout << "you have not created this grid number yet" << std::endl; 

    Grid dummy; 
    return dummy; 
} 

int Grid::create_swarm(std::string name) 
{ 
    Swarm new_swarm; 
    new_swarm.set_name("Protoswarm"); 
    swarmlist.push_back(new_swarm); 
    this->amount_of_swarms ++; 
    std::cout << "amount_of_swarms = " << amount_of_swarms << std::endl; 
    return 0; 
} 

Swarm Grid::get_swarm(int swarm_no) 
{ 
    std::cout << "grid.get_swarm(" << swarm_no << ") called." << std::endl; 
    if (swarm_no <= amount_of_swarms) 
     return swarmlist[swarm_no]; 
    else 
     std::cout << "oh oh - you have not this swarm in here..." << std::endl; 

    Swarm dummy; 
    return dummy; 
} 


using namespace std; 
int main(int argc, char* argv[]) 
{ 
    Sim mysim; 
    mysim.create_grid(); 
    mysim.get_grid(0).create_swarm("Alpha-Swarm"); 
    mysim.get_grid(0).create_swarm("Betaa-Swarm"); //doesn't work 

    Grid newgrid; 
    newgrid.create_swarm("Gamma-Swarm"); 
    newgrid.create_swarm("Delta-Swarm"); // works, but is not needed. 

    return 0; 
} 

Answer 1

Grid Sim::get_grid(int grid_no) {...} 

您是通过值返回，而不是通过引用。这意味着您要返回的是实际成员的副本。但是，对于您的情况，您希望通过引用返回，以便能够更改原始对象。您的代码将成为

Grid& Sim::get_grid(int grid_no) {...} 

请记住，但是，您将无法返回任何临时对象这种方式（如您的dummy网格），所以你需要改变你的方法来解决这个问题。如果你不想这样做，你仍然可以返回一个指针，虽然这会稍微改变语法。

Answer 2

你get_grid和get_swarm方法返回原始数组项目的副本。您应该将参考（或指针）返回至Grid或Swarm。