Overloaded function template never called

Question

I love templates, at least I would if I would understand them ;-). I implemented an overloaded operator using templates. I am now trying to specialise the function calls.

Here is what I do:

class Terminallog {
public:

    Terminallog();
    Terminallog(int);
    virtual ~Terminallog();

    template <class T>
    Terminallog & operator<<(const T &v);
    template <class T>
    Terminallog & operator<<(const std::vector<T> &v);
    template <class T>
    Terminallog & operator<<(const std::vector<T> *v);
    template <class T, size_t n>
    Terminallog & operator<<(const T(&v)[n]);
    Terminallog & operator<<(std::ostream&(*f)(std::ostream&));
    Terminallog & operator<<(const char v[]);

    //stripped code

};

//stripped code

template <class T>
Terminallog &Terminallog::operator<<(const T &v) {
    if (this->lineendet == true) {
        this->indent();
    }
    this->lineendet = false;
    std::cout << v;
    return *this;
}

template <class T>
Terminallog &Terminallog::operator<<(const std::vector<T> &v) {
    for (unsigned int i = 0; i < v.size(); i++) {
        std::cout << std::endl;
        this->indent();
        std::cout << i << ": " << v.at(i);
    }
    std::cout << std::flush;
    return *this;
}

template <class T>
Terminallog &Terminallog::operator<<(const std::vector<T> *v) {
    for (unsigned int i = 0; i < v->size(); i++) {
        std::cout << std::endl;
        this->indent();
        std::cout << i << ": " << v->at(i);
    }
    std::cout << std::flush;
    return *this;
}

template <class T, size_t n>
Terminallog &Terminallog::operator<<(const T(&v)[n]) {
    unsigned int elements = sizeof (v) / sizeof (v[0]);
    for (unsigned int i = 0; i < elements; i++) {
        std::cout << std::endl;
        this->indent();
        std::cout << i << ": " << v[i];
    }
    std::cout << std::flush;
    return *this;
}

inline
Terminallog &Terminallog::operator<<(std::ostream&(*f)(std::ostream&)) {
    if (f == static_cast<std::ostream & (*)(std::ostream&)> (std::endl)) {
        this->lineendet = true;
    }
    std::cout << f;
    return *this;
}

inline
Terminallog &Terminallog::operator<<(const char v[]) {
    if (this->lineendet == true) {
        std::cout << std::endl;
        this->indent();
        std::cout << v;
    }
    this->lineendet = false;
    std::cout << v;
    return *this;
}

//sripped code

Now I am trying something like

vector<int> *test3 = new vector<int>;
    test3->push_back(1);
    test3->push_back(2);
    test3->push_back(3);
    test3->push_back(4);

Terminallog clog(3);
clog << test3;

which compiles just fine. However executing the code, it prints the address of test3, instead all the elements. I conclude that the compiler thinks, that

Terminallog & operator<<(const T &v);

is a better match. However I don't know what to do about it. Where is the mistake in my code? Why is

Terminallog & operator<<(const std::vector<T> *v);

never called?

Answer 1

The type of test3 in your code is std::vector<int> *, but there is no Terminallog::operator<< that takes an argument of that type. There is one that takes const std::vector<int> * and it would be called if you did

clog << const_cast<const vector<int>*>(test3);

Incidentally, new'ing a vector is almost never a good idea.

Answer 2

For it to match the pointer-to-const-vector version, a const needs to be added to the type and this is not a top-level const (const-pointer-to-vector, i.e std::vector<T>* const). The chosen overload does not require any type conversions and is a better match.

I'd suggest completely removing the pointer overload, especially seeing that it just duplicates the vector reference overload. Instead just dereference the pointer.

clog << *test3;