Why standard containers use function templates instead of non-template Koenig operators

Question

This question is inspired by Issue with std::reference_wrapper. Let\' say, for example, operator< for std::vector. It\'s defined as a function t

Answer 1

Consider this code (A.h):

template <class T>
class A {
  public:
  T m_x;

  friend bool operator<(const A & lhs, const A & rhs) {
    return lhs.m_x < rhs.m_x;
  }
};

And main.cpp:

#include "A.h"

namespace buddy {
bool operator<(const A<double> & lhs, const A<double> &rhs) {
    return lhs.m_x > rhs.m_x;
};
}
using namespace buddy;
int main(int argc, char ** argv) {

  A<double> a1;
  A<double> a2;

  a1 < a2;

  return 0;
}

This code does not compile:

main.cpp:14:5: error: ambiguous overload for ‘operator<’ (operand types are ‘A’ and ‘A’) a1 < a2;

The reason of course is that both of the operator<'s are exact matches. On the other hand, if we change the first operator< to (defined outside the class):

template <class T>
bool operator<(const A<T> & lhs, const A<T> & rhs) {
  return lhs.m_x < rhs.m_x;
}

The compiler stops complaining: it's now a competition between an exact match, and a function template, so the exact match is used.

If operator< was defined in the fashion that you're suggesting, there would be no reasonable way for users of std::vector to redefine the behavior of operator<, short of specializing std::vector themselves, which is a lot more work.

In conclusion, the standard writers elected to make it easier to overload operator<, than to provide an operator< that might be more useful in certain situations. I think they made the right choice.