Variadic Templates: Overload resolution

Question

#include <iostream>

template<typename T>
bool pair_comparer(T a, T b) {
  return a != b;
}

template<typename T, typename... Args>
bool pair_comparer(T a, T b, Args... args) {
  return a == b && pair_comparer(args...);
}

int main() {
    bool areSame = pair_comparer(1, 1, 2, 2, 6, 6);
    std::cout << "areSame " << areSame << endl;    // prints 0
    return 0;
}

I don't expect the example to compile , but it is . How is the call to pair_comparer with two arguments resolved here ? Is there something I am missing

Answer 1

In the first call to pair_comparer you have:

a = 1 , b = 1, and args = [2, 2, 6, 6]

and then you have a recursive call to pair_comparer and this time you have:

a = 2 , b = 2, and args = [6, 6]

finaly you have a call to pair_comparer(T a, T b) with: args = [6, 6]

But you have to put return a == b; instead of return a != b; in bool pair_comparer(T a, T b) to get the right answer. it means all the numbers are equal by pair

But if you start to call by two arguments only:

For example with:

bool areSame = pair_comparer(6, 6);

The program will call directly: pair_comparer(T a, T b) because it is the first match (we have only 2 parameters and not more)