C ++成员函数指针 [英] C++ Member Function Pointers

问题描述

我在C ++做一个小游戏，我发现类成员函数指针。
我不想让他们以正确的方式工作，但这里是我的尝试。

  //一个结构体，其中的函数指针将被存储为调用
 //顺便说一句，有一种方法来做类同样的事情吗？ 
 //或者在C ++中结构仍然很好吗？ （感觉像使用char而不是字符串）
 typedef struct s_dEntitySpawn 
 {
 std :: string name; 
 void（dEntity :: * ptr）（）; 
} t_dEntitySpawn 
 
 //填充结构体，如果实体的类名是actor_basicnpc，
 //然后我想做一个调用像ent-> spawnBasicNPC 
 t_dEntitySpawn dEntitySpawns [ ] = {
 {actor_basicnpc，& dEntity :: spawnBasicNPC}，
 {0，0} 
}; 
 
 //这是每个实体被分析的地方
 //和我调用函数指针
 void dEntitiesManager :: spawnEntities（）
 {
实体* 
 t_dEntitySpawn * spawn; 
 
 [...] 
 
 //这里出错，对我来说很奇怪
 if（！spawn-> name.compare - > getClassname（）））
 ent-> * spawn.ptr（）; 
 
 [...] 
} 
  

解决方案

/ div>

我认为你正在寻找的是

 （ent-> *（spawn- > ptr））（）; 
  

让我们剖析一下。首先，我们需要得到实际的成员函数指针，它是

  spawn-> ptr 
  

由于 spawn 是一个指针，我们必须使用 - > 以选择 ptr 字段。

一旦我们这样做，我们需要使用指针到成员选择运算符来告诉 ent 来选择适当的成员函数：

  ent-> *（spawn-> ptr）
  

b $ b

最后，要调用函数，我们需要告诉C ++调用这个成员函数。由于C ++中的操作符优先级问题，您首先必须对整个表达式进行括号，以评估成员函数，因此我们有

  （ent-> *（spawn-> ptr））（）; 
  

这是值得的，这是最古怪的C ++代码之一，我见过一会儿。 ： - ）

在一个完全无关的笔记上，因为你使用的是C ++，我会避免使用 typedef struct 。只需说

  struct t_dEntitySpawn {
 std :: string name; 
 void（dEntity :: * ptr）（）; 
}; 
  

希望这有助于！

I'm doing a little game in C++ and I'm discovering the class members function pointers. I don't have any idea to make them work in the right way, but here is my attempt.

// A struct where the function pointer will be stored for the call
// By the way, is there a way to do the same thing with classes ?
// or are structs still fine in C++ ? (Feels like using char instead of string)
typedef struct      s_dEntitySpawn
{
  std::string       name;
  void          (dEntity::*ptr)();
} t_dEntitySpawn;

// Filling the struct, if the entity's classname is "actor_basicnpc",
// then I would like to do a call like ent->spawnBasicNPC
t_dEntitySpawn      dEntitySpawns[] = {
  { "actor_basicnpc", &dEntity::spawnBasicNPC },
  { 0, 0 }
};

// This is where each entity is analyzed
// and where I call the function pointer
void            dEntitiesManager::spawnEntities()
{
  dEntity       *ent;
  t_dEntitySpawn    *spawn;

    [...]

      // It makes an error here, feels very weird for me
      if (!spawn->name.compare(ent->getClassname()))
        ent->*spawn.ptr();

    [...]
}

Could you please give me a good advice about the right way to implement them ?

Best regards.

解决方案

I think that the line you're looking for is

(ent->*(spawn->ptr))();

Let's dissect this. First, we need to get to the actual member function pointer, which is

spawn->ptr

Since, here, spawn is a pointer, and we have to use -> to select the ptr field.

Once we have that, we need to use the pointer-to-member-selection operator to tell ent to select the appropriate member function:

ent->*(spawn->ptr)

Finally, to call the function, we need to tell C++ to invoke this member function. Due to operator precedence issues in C++, you first have to parenthesize the entire expression that evaluates to the member function, so we have

(ent->*(spawn->ptr))();

For what it's worth, this is one of the weirdest lines of C++ code that I've seen in a while. :-)

On a totally unrelated note, because you're using C++, I would avoid using typedef struct. Just say

struct t_dEntitySpawn {
  std::string name;
  void (dEntity::*ptr)();
};

Hope this helps!

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