How to use source_location in a variadic template function?

Question

The C++20 feature std::source_location is used to capture information about the context in which a function is called. When I try to use it with a variadic tem

Answer 1

The first form can be made to work, by adding a deduction guide:

template <typename... Ts>
struct debug
{    
    debug(Ts&&... ts, const std::source_location& loc = std::source_location::current());
};

template <typename... Ts>
debug(Ts&&...) -> debug<Ts...>;

Test:

int main()
{
    debug(5, 'A', 3.14f, "foo");
}

DEMO

Answer 2

Just put your arguments in a tuple, no macro needed.

#include <source_location>
#include <tuple>

template <typename... Args>
void debug(
    std::tuple<Args...> args,
    const std::source_location& loc = std::source_location::current())
{
    std::cout 
        << "debug() called from source location "
        << loc.file_name() << ":" << loc.line()  << '\n';
}

And this works^*.

Technically you could just write:

template <typename T>
void debug(
    T arg, 
    const std::source_location& loc = std::source_location::current())
{
    std::cout 
        << "debug() called from source location "
        << loc.file_name() << ":" << loc.line()  << '\n';
}

but then you'd probably have to jump through some hoops to get the argument types.

---

_{* In the linked-to example, I'm using <experimental/source_location> because that's what compilers accept right now. Also, I added some code for printing the argument tuple.}

Answer 3

template <typename... Args>
void debug(Args&&... args,
           const std::source_location& loc = std::source_location::current());

"works", but requires to specify template arguments as there are non deducible as there are not last:

debug<int>(42);

Demo

Possible (not perfect) alternatives include:

• use overloads with hard coded limit (old possible way to "handle" variadic):

  // 0 arguments
  void debug(const std::source_location& loc = std::source_location::current());
  
  // 1 argument
  template <typename T0>
  void debug(T0&& t0,
             const std::source_location& loc = std::source_location::current());
  
  // 2 arguments
  template <typename T0, typename T1>
  void debug(T0&& t0, T1&& t1,
             const std::source_location& loc = std::source_location::current());
  
  // ...
  

  Demo

• to put source_location at first position, without default:

  template <typename... Args>
  void debug(const std::source_location& loc, Args&&... args);
  

  and

  debug(std::source_location::current(), 42);
  

  Demo

• similarly to overloads, but just use tuple as group

  template <typename Tuple>
  void debug(Tuple&& t,
             const std::source_location& loc = std::source_location::current());
  

  or

  template <typename ... Ts>
  void debug(const std::tuple<Ts...>& t,
             const std::source_location& loc = std::source_location::current());
  

  with usage

  debug(std::make_tuple(42));
  

  Demo

Answer 4

Not a great solution but... what about place the variadic arguments in a std::tuple?

I mean... something as

template <typename... Args>
void debug (std::tuple<Args...> && t_args,
            std::source_location const & loc = std::source_location::current());

Unfortunately, this way you have to explicitly call std::make_tuple calling it

debug(std::make_tuple(1, 2l, 3ll));