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Home/ Questions/Q 7428699
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T08:50:43+00:00 2026-05-29T08:50:43+00:00

I’m doing a little game in C++ and I’m discovering the class members function

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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.

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  1. Editorial Team
    Editorial Team
    2026-05-29T08:50:45+00:00Added an answer on May 29, 2026 at 8:50 am

    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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