C++ dereferencing an instance - c++

Question

I'm new to C++ and I'm trying to dry up my code, for example:
void gethit () {
Gert.hitbox(AA.x, AA.damage);
Gert.hitbox(AB.x, AB.damage);
Gert.hitbox(AC.x, AC.damage);
Gert.hitbox(AD.x, AD.damage);
Gert.hitbox(Terbil.x, Terbil.damage);
}
AA, AB, AC, AD and Terbil are all instances of a class called Entity with variables x and damage.
Every time I want to add a new instance I'll have to come into this function and add it manually. I'm trying to add all of the address of the instances to an array like so:
void * p_enemys[10];
p_enemys[0] = &AA;
p_enemys[1] = &AB;
p_enemys[2] = ∾
p_enemys[3] = &AD;
p_enemys[4] = &Terbil;
Just wondering how I could call a function from the instance via the array, I tried to do
for(int i = 0; i < 10; i++;) {
Gert.hitbox(p_enemys[i].x, p_enemys[i].damage);
}
and g++ compiler spits out: "request for member `damage' in `p_enemys[i]', which is of non-aggregate type `void *'"
I don't really need to use arrays specifically any help is very appreciated.
Changes made, thanks #gldraphael!
vector <Entity*> p_Enemys(10);
void gethit () {
for (int i = 0; i < 10; ++i) {
Entity * ienemy = (Entity*) p_Enemys[i];
Gert.hitbox((ienemy->x), (ienemy->damage));
}
}

Answer 1

You can make an a std::vector as follows:
std::vector <Entity*> p_Enemys(10);
The assigning part remains the same:
p_enemys[0] = &AA;
p_enemys[1] = &AB;
p_enemys[2] = &AC;
p_enemys[3] = &AD;
p_enemys[4] = &Terbil;
You can then loop through the p_enemys as follows:
for(auto i : p_enemys) {
Gert.hitbox(i->x, i->damage);
}
So what was it that you missed?
The array was declared as an array of void*
So, in the loop, p_enemys[i] returned a void*.
Also class/struct members are accessed using a dereferencing operator ->. You used the membership operator . instead.
So this code should have worked instead:
for(int i = 0; i < p_enemys.size(); i++;) { //
Entity * ienemy = (Entity*) p_enemys[i]; // cast the void* into a Entity*
Gert.hitbox(ienemy->x, ienemy->damage);
}
As a general rule, avoid void*s whenever possible.