Using fold expressions to print all variadic arguments with newlines inbetween

Question

The classic example for C++17 fold expressions is printing all arguments:

template
void print(Args ... args)
{
    (cout << ... &l         


        

Answer 1

repeat takes a function object f, and return a new function object. The return value runs f on each of its args. It "repeats" f on each of its args.

template
auto repeat( F&& f ) {
  return [f=std::forward(f)](auto&&...args)mutable{
    ( void(f(args)), ... );
  };
}

Use:

repeat
( [](auto&&x){ std::cout << x << "\n"; } )
( args... );

This uses fold expressions, but only indirectly. And honestly, you could have written this in C++14 (just the body of repeat would be uglier).

We could also write a streamer that works with << to do it "more inline" and use fold expressions directly:

template
struct ostreamer_t {
  F f;
  friend std::ostream& operator<<( std::ostream& os, ostreamer_t&& self ) {
    std::move(self).f(os);
    return os;
  }
};

template
ostreamer_t ostreamer( F&& f ) { return {std::forward(f)}; }

then we use it like this:

(std::cout << ... << ostreamer([&](auto&& os){ os << " " << args;}));

ostreamer takes a function object. It returns an object that overloads << such that when you pass it an ostream on the left, it invokes the function object with the ostream.

No temporary stream is created.

Live examples.