c++ - Using fold expressions to print all variadic arguments with newlines inbetween -

the classic example c++17 fold expressions printing arguments:

template<typename ... args> void print(args ... args) {     (cout << ... << args); } 

example:

print("hello", 12, 234.3, complex<float>{12.3f, 32.8f}); 

output:

hello12234.3(12.3,32.8) 

i'd add newlines output. however, can't find way that, best i've found far:

template<typename ... args> void print(args ... args) {     (cout << ... << ((std::ostringstream{} << args << "\n").str())); } 

this isn't zero-overhead, constructs temporary ostringstream each argument.

the following versions don't work either:

(cout << ... << " " << args);  error: expression not permitted operand of fold expression 

and

(cout << ... << (" " << args));  error: invalid operands binary expression  

i understand why last 2 versions don't work. there more elegant solution problem, using fold expressions?

repeat takes function object f, , return new function object. return value runs f on each of args. "repeats" f on each of args.

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

use:

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

this uses fold expressions, indirectly. , honestly, have written in c++14 (just body of repeat uglier).

we write streamer works << "more inline" , use fold expressions directly:

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

then use this:

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

ostreamer takes function object. returns object overloads << such when pass ostream on left, invokes function object ostream.

no temporary stream created.

live examples.