I have written a class for generating debug messages as console output like so.

class mDebug 
{
    std::ostringstream logStream;

public:

    mDebug();
    mDebug(const mDebug& other);
    ~mDebug();

    template <typename T>
    mDebug& operator<<(const T& value) {logStream << value;}
};

It contains a templated catchall << operator that allows me to write neat debug output of primitive types like so.

  mDebug() << 1.2 << "hello";

True debugging power unfolds when I implement global debug operators for the more complex objects I have in my framework, for example for a Vector:

mDebug operator<<(mDebug dbg, const Vector<T> &o)
{
    dbg << "sz:" << o.size() << " | ";
    dbg << "[";
    if (o.size() > 1)
        for (uint i = 0; i < o.size()-1; i++)
            dbg << o[i] << ",";
    if (o.size() > 0)
        dbg << o[o.size()-1];
    dbg << "]";
    return dbg;
}

and then I can

Vector<double> vec;
vec << 1.2 << 1.3 << 1.4;
mDebug() << vec;

Cool. The one thing left that doesn’t let me sleep at night is that the global mDebug Vector operator takes and returns a copy of the debug stream rather than a reference. I would much prefer to have it like so

mDebug& operator<<(mDebug& dbg, const Vector<T> &o)

but then a compilation error happens that says:

error: no match for ‘operator<<’ (operand types are ‘std::ostringstream’ {aka ‘std::__cxx11::basic_ostringstream’} and ‘const Vector’)
30 | logStream << value;
| ~~~~~~~~~~^~~~~~~~

Apparently, the member operator mDebug& operator<<(const T& value) has taken precedence over the global mDebug operator<<(mDebug& dbg, const Vector<T> &o) operator and my “catchall” function now also catches the Vector and tries to stream it directly into logStream, which won’t work. I don’t really understand the order of things here. How can I accomplish that all my objects can be streamed into my mDedug class using global operators while passing references AND have the catchall function declared for all primitive types?