c + +使用一個類中的一類（一類），並調用其功能

Question

我有以下問題：我想從好老C項目重做C++，使一切class(y) :)並保持它從一開始可擴展性。

它是細胞的網格上的一個模擬（A羣被部分），所以決定以下結構：

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

我在單獨的頭文件中定義的類。那麼當然，我需要能夠在grid，swarm和cell中調用函數。我想要做的直截了當：

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

與電網作爲返回參數

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; 
} 

它調用的函數，並且只要在電網未進行任何更改的作品。這些似乎在太空中消失了。可能是一個指針錯誤，但我不能找到一個錯誤，因爲完全相同的代碼工作的Simulation類...

更多來源：

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; 
} 

我可以隨時調用create_swarm功能我想要，但是羣體永遠不會出現，並且櫃檯不會在該網格中提升，只要功能在那裏即可。我錯過了什麼嗎？它真的只是一個指針錯誤？爲什麼這段代碼工作，如果我這樣稱呼它：

Grid newgrid; 
newgrid.create_swarm(); 

一個快速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) {...} 

您是通過值返回，而不是通過引用。這意味着您要返回的是實際成員的副本。但是，對於您的情況，您希望通過引用返回，以便能夠更改原始對象。您的代碼將成爲

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

請記住，但是，您將無法返回任何臨時對象這種方式（如您的dummy網格），所以你需要改變你的方法來解決這個問題。如果你不想這樣做，你仍然可以返回一個指針，雖然這會稍微改變語法。

Answer 2

你get_grid和get_swarm方法返回原始數組項目的副本。您應該將參考（或指針）返回至Grid或Swarm。