如果可变参数模板中有这样的function？

Question

for example there is such a template

template<class ... Types> void f(Types ... args);

f();       // OK: args contains no arguments
f(1);      // OK: args contains one argument: int
f(2, 1.0); // OK: args contains two arguments: int and double

and I want to do so in it

template<class ... T>
void f2(T... args) {
  // option 1
  // > hello // world // human
  std::cout <<(args / ...);

  // option 2
  // > hello // world
  std::cout <<((args / ...) and -1 args); // ../
}

in the example, option 2 We are concatenating the Hello and world string and at the same time not using the human , since we don't need it yet.

if it is possible of course

f2("hello", "world", "human");

then

get one less argument inside the function to use it inside the function

For this call:

f2("A", "b", 12);

Expected result should be equivalent to this:

std::cout << "A";
std::cout << "b";

// no statement or different action for: std::cout << 12;

and if possible without arrays and recursions if there is no such functionality, then write that it is not there.

Answer 1

template<bool condition, typename T>
void printIf(T&& arg)
{
    if constexpr (condition) std::cout << arg;
}

template<size_t...indexes, class ... T>
void printAllButLastHelper(std::integer_sequence<size_t, indexes...> v, T&&... args) {
    (printIf<sizeof...(args) - 1 != indexes>(args), ...);
}

template<class ... T>
void f2(T&&... args) {
    std::cout << __PRETTY_FUNCTION__ << '\n';
    printAllButLastHelper(
        std::make_integer_sequence<size_t, sizeof...(args)>{}, 
        std::forward<T>(args)...);
    std::cout << '\n';
}

https://godbolt.org/z/3dojcd