Is it possible to have an “auto” member variable?

Question

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

Answer 1

This is what the C++ draft standard has to say about using auto for member variables, in section 7.1.6.4 auto specifier paragraph 4:

The auto type-specifier can also be used in declaring a variable in the condition of a selection statement (6.4) or an iteration statement (6.5), in the type-specifier-seq in the new-type-id or type-id of a new-expression (5.3.4), in a for-range-declaration, and in declaring a static data member with a brace-or-equal-initializer that appears within the member-specification of a class definition (9.4.2).

Since it must be initialized this also means that it must be const. So something like the following will work:

struct Timer
{
  const static int start = 1;
}; 

I don't think that gets you too much though. Using template as Mark suggests or now that I think about it some more maybe you just need a variant type. In that case you should check out Boost.Variant or Boost.Any.

Answer 2

Indirectly, provided that you don't reference a member of the class.

This can also now be achieved through deduction guides, these were introduced in C++17 and have recently (finally) support in VC++ has been added (clang and GCC already had it).

https://en.cppreference.com/w/cpp/language/class_template_argument_deduction

For example:

template <typename>
struct CString;

template <typename T, unsigned N>
struct CString<std::array<T, N>>
{
    std::array<T, N> const Label;

    CString(std::array<T, N> const & pInput) : Label(pInput) {}
};

template <typename T, std::size_t N>
CString(std::array<T, N> const & pInput) -> CString<std::array<T, N>>;

https://godbolt.org/z/LyL7UW

This can be used to deduce class member types in a similar manner to auto. Although the member variables need to be Dependant somehow on the constructor arguments.

Answer 3

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.

Answer 4

No. Each constructor could have its own initializer for start, so there could be no consistent type to use.

If you do have a usable expression, you can use that:

struct Timer {

   Foo getAFoo();

   delctype(Timer().getAFoo().Bar()) start;

   Timer() : start(getAFoo().Bar()) { /***/ }
};