Sorting an STL Vector of pointers

Hi all,

I have a stl::vector containing pointers to objects. I need to sort this vector based on a member variable (accessible by a getter function). Can someone give me a clue as to how to go about this?

I've tried a few things - but nothing has really worked, I think mainly because every bit of information I can find on this doesn't talk about pointers...

For instance, one of the methods I tried was:

Code:

---


class Ccharacter {
  int m_spawnOrder;

  int getSpawnOrder();

public:
    Ccharacter(int spawnOrder);
    bool operator<(const Ccharacter& a) const;

};

Ccharacter::Ccharacter(int spawnOrder) {
    m_spawnOrder = spawnOrder
}

int Ccharacter::getSpawnOrder() {
    return m_spawnOrder;
}

bool Ccharacter::operator<(const Ccharacter& a) const {
 
  return this->getSpawnOrder() < a.getSpawnOrder();
}

---

I have a vector like such that I need to sort:

Code:

---


vector <Ccharacter *> listOfCharacters;

---

The operator< is not getting called at all in this scenario.

Cheers,

B

Of course not. You're comparing Ccharacter*, which has its own < operator.

You need to tell sort not to use the operator at all, but instead your own comparison predicate.

Code:

---

struct compareCharacters {
  bool operator ()(Ccharacter *lhs, Ccharacter *rhs) { return *lhs < *rhs; }
};

std::sort(listOfCharacters.begin(), listOfCharacters.end(), compareCharacters());

---

There are more generic ways of solving this problem, but this will do for now.

Exceptional thank you! Although - i'm not 100% sure why it worked - would someone be so kind as to give me a quick explanation as to why I need the the struct to do the test? I thought sort would use the < if no function was passed to it...

But again thanks ;)

It does, but that's the problem. For Ccharacter *lhs, Ccharacter *rhs, it does lhs < rhs, but you want *lhs < *rhs. Thus the struct (called a "functor" or "function object", because it overloads operator ()) passed as the third argument.

OH! I see - that makes perfect sense! Thanks very much for the help. :)