c++ - lua函数到C++函数

Question

我正在努力将 lua 程序转换为 C++ 程序，但我遇到了障碍，我不知道如何将其转换为 C++

function newPool()
    local pool = {}
    pool.species = {} --imports data from local species = {}
    pool.generation = 0
    pool.innovation = Outputs
    pool.currentSpecies = 1
    pool.currentGenome = 1
    pool.currentFrame = 0
    pool.maxFitness = 0

    return pool
end

我知道这两种语言的许多基础知识，我知道它在 lua 中有效，但我需要它在 C++ 中。谁能帮帮我？

Answer 1

Lua 有一个叫做 Tables 的东西它允许您在没有预定义 struct 的情况下添加键值对就像在 C/C++ 中一样。因此，您发布的 Lua 代码是将键值对添加到 pool (阅读代码中的注释):

local pool = {}           -- Declare a new Table
pool.species = {}         -- Add another Table to pool called 'species'
pool.generation = 0       -- Add the key 'generation' with value '0'
pool.innovation = Outputs -- Add the key 'innovation' with value 'Outputs'
pool.currentSpecies = 1   -- Add the key 'currentSpecies' with value '1'
pool.currentGenome = 1    -- Add the key 'currentGenome' with value '1'
pool.currentFrame = 0     -- Add the key 'currentFrame' with value '0'
pool.maxFitness = 0       -- Add the key 'maxFitness' with value '0'

在 C++ 中，您有多种选择。 1)你可以创建一个struct并声明你需要什么(我猜测某些数据类型，但如果你有完整的 Lua 程序，你可以弄清楚它们):

struct Pool
{
   Species species;     // You'll have to define Species in another struct
   int generation;
   SomeEnum innovation; // You'll have to define SomeEnum in an enum
   int currentSpecies;
   int currentGenome;
   int currentFrame;
   int maxFitness;
}

如果你有一个类，那么你可以使用 struct Pool如下所示(将上面的 struct Pool 定义添加到 class Kingdom 上方的 .h 文件中):

// I'm doing this as a class since you are programming in C++ and I
// assume you will want to add more functions to operate on similar
// objects.
class Kingdom
{
public:
   Kingdom();
   Pool* NewPool();
private:
   Pool _pool;
}

在您的 .cpp 文件中:

#include "Kingdom.h"

Kingdom::Kingdom()
{
  // _pool.species = whatever you define struct Species as
  _pool.generation = 0;
  _pool.innovation = SomeEnum::Outputs; // You'll have to define SomeEnum
  _pool.currentSpecies = 1;
  _pool.currentGenome = 1;
  _pool.currentFrame = 0;
  _pool.maxFitness = 0; 
}

Pool* Kingdom::NewPool()
{
  Pool* newPool = new Pool;
  memcpy(newPool, &_pool, sizeof(Pool)); // Make a copy
  return newPool;                        // Return the new copy

  // The newPool value is dynamic memory so when the calling function is done
  // with newPool it should delete it, example:
  // Kingdom myKingdom;
  // Pool* myNewPoolStruct = myKingdom.NewPool();
  // ... do some coding here
  // delete myNewPoolStruct;
}

选项 2) 如果您所有的键值对都是同一类型；即所有键都是 std::string所有值都是 int .请记住，Lua 代码使用表，因此您可以在 C++ 中使用 std::map<> 创建等效项.然后你可以使用 std::map<std::string, int>如下:

// In your .h file change
Pool* NewPool();
Pool _pool;
// to
std::map<std::string, int> NewPool();
std::map<std::string, int> _pool;

然后在您的 .cpp 文件中将构造函数更改为:

Kingdom::Kingdom()
{
  _pool["species"] = 0;    // Some int representation of species
  _pool["generation"] = 0;
  _pool["innovation"] = 1; // Some int representation of Outputs
  _pool["currentSpecies"] = 1;
  _pool["currentGenome"] = 1;
  _pool["currentFrame"] = 0;
  _pool["maxFitness"] = 0;
}

std::map<std::string, int> NewPool()
{
  std::map<std::string, int> newPool;
  newPool = _pool;  // Copy - double check this against std::map
  return newPool;   // Double check this is a true copy and not a pointer
}

与 std::map您可以像您提供的 Lua 代码一样即时创建键值对。简而言之，我会选择 struct Pool方法是因为 std::map<>你必须记住字符串，这不是好的做法，你的 IDE 应该有智能感知，它将始终向你显示 struct Pool 的内容。每当你点击 .或 ->运营商。