c++ - How to write a function wrapper for cout that allows for expressive syntax? -

i'd wrap std::cout formatting, so:

mycout([what type?] x, [optional args]) {     ... // formatting on x first     std::cout << x; } 

and still able use expressive syntax like

mycout("test" << << endl << somevar, indent) 

instead of being forced more verbose like

mycout(std::stringstream("test") << ...) 

how can implement this? type make x?

edit: added consideration optional arguments

how this:

struct mycout {};  extern mycout mycout;  template <typename t> mycout& operator<< (mycout &s, const t &x) {   //format x please   std::cout << x;   return s; } 

and put mycout mycout; 1 .cpp file.

you can use mycout this:

mycout << "test" << x << std::endl; 

and call template operator<< can formatting.

of course, can provide overloads of operator special formatting of specific types if want to.

edit

apparently (thanks @soon), standard manipulators work, few more overloads necessary:

mycout& operator<< (mycout &s, std::ostream& (*f)(std::ostream &)) {   f(std::cout);   return s; }  mycout& operator<< (mycout &s, std::ostream& (*f)(std::ios &)) {   f(std::cout);   return s; }  mycout& operator<< (mycout &s, std::ostream& (*f)(std::ios_base &)) {   f(std::cout);   return s; } 

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edit 2

i may have misunderstoor original requirements slightly. how (plus same manipulator overloads above):

struct mycout {   std::stringstream s;    template <typename t>   mycout& operator << (const t &x) {     s << x;     return *this;   }    ~mycout() {     somehow_format(s);     std::cout << s.str();   } };  int main() {   double y = 1.5;   mycout() << "test" << y; }