Question

I am wondering. I hit a problem and here is a small repoduce. Essentially i want to forward everything. The problem is, using the first << will cause an error with o<<1 (or o<<SomeUserStruct(). If i include the second i get errors about it being ambiguous. Is there a way i can write this code so it will use T& when it can otherwise uses T?

#include <iostream>
struct FowardIt{
    template<typename T> FowardIt& operator<<(T&t) { std::cout<<t; return *this; }
    //template<typename T> FowardIt& operator<<(T t) { std::cout<<t; return *this; }
};

struct SomeUserStruct{};

int main() {
    FowardIt o;
    o << "Hello";
    int i=1;
    o << i;
    o << 1;
    o << SomeUserStruct();
}

Answer 1

template<typename T> FowardIt& operator<<(const T&t)
                                        //^^^^^ put const here

Make the parameter const reference, as shown above. Because temporaries cannot be bound to non-const reference. You don’t need to define another function. Just make the parameter const, the problem will be solved.

It would also be better if you make the function template const as well by putting const to the rightmost side of the function as:

template<typename T> 
const FowardIt& operator<<(const T&t) const
^^^^^                      ^^^^^      ^^^^^
  |                          |          |
  |                          |          put const here as well
  |                          put const here
  |
  You've to make the return-type also const
  since it can't return non-const reference anymore

If you do so, then you can call this function on const objects as well:

void f(const FowardIt &o)//note: inside the function, o is an const object!
{
    o << 1;
    o << SomeUserStruct();
}