overloading operators c++

Define a custom comparator function or functor:

struct foo {   //....   bool operator<(const foo& f) const { .... }};inline bool compare_foos(const foo* const lhs,                         const foo* const rhs) {   return *lhs < *rhs;}....typedef std::priority_queue< foo*,                              std::vector<foo*>,                             compare_foos > foo_queue;

Quote:

Original post by snk_kid Define a custom comparator function or functor: struct foo { //.... bool operator<(const foo& f) const { .... }};inline bool compare_foos(const foo* const lhs, const foo* const rhs) { return *lhs < *rhs;}....typedef std::priority_queue< foo*, std::vector<foo*>, compare_foos > foo_queue;

Or, if you're really slick, you can generalize it:

template<class T, class Compare = std::less<T> > struct pointeeCompare{  bool operator()(T* lhs, T* rhs)  {     return Compare()(*lhs, *rhs);  }};....typedef std::priority_queue< foo*,                              std::vector<foo*>,                             pointeeCompare<foo> > foo_queue;

Quote:

Original post by Couvillion Or, if you're really slick, you can generalize it

And if you're really, really slick, you'll make it an adaptable functor. [grin]

template<class T, class Compare = std::less<T> > struct pointeeCompare :   std::binary_function<T,T,bool>{  ...}

Quote:

Original post by Couvillion Or, if you're really slick, you can generalize it: template<class T, class Compare = std::less<T> > struct pointeeCompare{ bool operator()(T* lhs, T* rhs) { return Compare()(*lhs, *rhs); }};....typedef std::priority_queue< foo*, std::vector<foo*>, pointeeCompare<foo> > foo_queue;

Not quite, that code will bork if you pass the address of a free-function there-for its not generalized that much and what happened to constant correctness [grin]

By the way the reason why i showed only a simple function was i'm assuming that he probably has never come across things like functors before otherwise he would have never asked that question in the first place.