Is it possible to have an "auto" member variable?

Oleksiy picture Oleksiy · Aug 13, 2013 · Viewed 24.2k times · Source

For example I wanted to have a variable of type auto because I'm not sure what type it will be.

When I try to declare it in class/struct declaration it's giving me this error:

Cannot deduce auto type. Initializer required

Is there a way around it?

struct Timer {

    auto start;

};

Answer

Mark Garcia picture Mark Garcia · Aug 13, 2013

You can, but you have to declare it static and const:

struct Timer {
    static const auto start = 0;
};

A working example in Coliru.

With this limitation, you therefore cannot have start as a non-static member, and cannot have different values in different objects.

If you want different types of start for different objects, better have your class as a template

template<typename T>
struct Timer {
    T start;
};

If you want to deduce the type of T, you can make a factory-like function that does the type deduction.

template<typename T>
Timer<typename std::decay<T>::type> MakeTimer(T&& startVal) {   // Forwards the parameter
   return Timer<typename std::decay<T>::type>{std::forward<T>(startVal)};
}

Live example.