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

Question

I\'d like to wrap std::cout for formatting, like so:

mycout([what type?] x, [optional args]) {
    ... // do some formatting on x first
    std:         


        

Answer 1

You can use this kind of class:

#include 

using namespace std;

class CustomOut
{
public:

    template
    CustomOut& operator<<(const T& obj)
    {
        cout << " my-cout " << obj;
        return *this;
    }

};

int main()
{
    CustomOut mycout;
    mycout << "test" << 4 << "\n" << 3.4;
}

You'd need more code to use std::endl and other functors, so i've used here simple \n instead.