Declare variables that depend on unknown type in template functions

Question

Suppose I'm writing a template function foo that has type parameter T. It gets an object of type T that must have method bar(). And inside foo I want to create a vector of objects of type returned by bar.

In GNU C++ I can write something like that:

template<typename T>
void foo(T x) {
    std::vector<__typeof(x.bar())> v;
    v.push_back(x.bar());
    v.push_back(x.bar());
    v.push_back(x.bar());
    std::cout << v.size() << std::endl;
}

How to do the same thing in Microsoft Visual C++? Is there some way to write this code that works in both GNU C++ and Visual C++?

Answer 1

You can do that in standard c++

template<typename T>
struct id { typedef T type; };

template<typename T>
id<T> make_id(T) { return id<T>(); }

struct any_type {
  template<typename T>
  operator id<T>() const { return id<T>(); }
};

template<typename T, typename U>
void doit(id<T>, U& x) {
  std::vector<T> v;
  v.push_back(x.bar());
  v.push_back(x.bar());
  v.push_back(x.bar());
  std::cout << v.size() << std::endl;
}

template<typename T>
void foo(T x) {
    doit(true ? any_type() : make_id(x.bar()), x);
}

See Conditional Love for an explanation.