Specialize function template with decltype trailing return type

Question

In C++11, how can I specialise a function template which is declared with a "complicated" trailing return type using decltype? The following works in GCC but produces "error C2912: explicit specialisation 'int f(void)' i s not a specialisation of a function template" in VC2013:

#include <iostream>

int myint() { return 1; }

template<class T>
auto f() -> decltype(myint()) // this seems to cause problems
{
    std::cout << "general\n";
    return 1;
}

template <>
auto f<double>() -> decltype(myint())
{
    std::cout << "special\n";
    return 2;
}

int main()
{
    f<int>();
    f<double>(); // compiler error in VC, but not in GCC
}

I say "complicated" in lack of a technically precise word because I'm not sure what makes the difference. For example, the following decltype, using a built-in operation not depending on any function result type, works fine with template specialisation:

auto f() -> decltype(1 + 1)

So, my questions (all related to each other):

Is my code correct C++11?
Is this a VC bug?
How could I ever specialise std::begin and std::end (and thus provide range-based for loops) for an unchangeable legacy container class if this kind of specialization does not work?

Ali Ali · Accepted Answer · 2014-02-15T11:47:20

Is my code correct C++11?

Looks correct to me. Also, compiles cleanly both with gcc and clang with -Wall -Wextra.

Is this a VC bug?

Most likely. VC is infamous in this respect, see for example What exactly is "broken" with Microsoft Visual C++'s two-phase template instantiation? or google msvc two-phase lookup.

How could I ever specialise std::begin and std::end (and thus provide range-based for loops) for an unchangeable legacy container class if this kind of specialization does not work?

For the code you provided, a workaround would be to use a typedef:

#include <iostream>

int myint() { return 1; }

typedef decltype(myint()) return_type;

template<class T>
return_type f()
{
    std::cout << "general\n";
    return 1;
}

template <>
return_type f<double>()
{
    std::cout << "special\n";
    return 2;
}

int main()
{
    f<int>();
    f<double>();
}

All three mainstream compilers (gcc, clang, vs) seem to be happy with this code.

UPDATE:

How could I ever specialise std::begin and std::end (and thus provide range-based for loops) for an unchangeable legacy container class if this kind of specialization does not work?
[And from the comments:] I thought specialising std::begin and std::end was always the best approach.

After giving it some thought, specializing std::begin() and std::end() would be my last resort. My first attempt would be to provide member begin() and end() functions; unfortunately, it is not an option for you because you cannot modify the corresponding code. Then, my second attempt would be to provide free functions in my own namespace:

#include <iostream>
#include <initializer_list>
#include <vector>

namespace my_namespace {

template <typename T> class my_container;
template <typename T> T* begin(my_container<T>& c);
template <typename T> T* end(my_container<T>& c);

template <typename T>
class my_container {

  public:

    explicit my_container(std::initializer_list<T> list) : v(list) { }

    friend T* begin<>(my_container& c);
    friend T* end<>(my_container& c);

  private:

    std::vector<T> v;
};

template <typename T>
T* begin(my_container<T>& c) {

  return c.v.data();
}

template <typename T>
T* end(my_container<T>& c) {

  return c.v.data()+c.v.size();
}

}

int main() {

  my_namespace::my_container<int> c{1, 2, 3};

  for (int i : c)
    std::cout << i << '\n';
}

This approach must work if you were able to specialize std::begin() and std::end() for the container. It also works if you do it in the global namespace (that is, you simply omit the namespace my_namespace { and closing }) but I prefer to put my implementation into my own namespace.

Specialize function template with decltype trailing return type

1 Answers