Visual studio - workaround error c2770 on std::enable_if

Question

taking the answer from here : iterate over tuple regarding printing std::tuple components, here is the code :

template<std::size_t I = 0, typename... Tp>
 typename std::enable_if<I == sizeof...(Tp), void>::type
print(const std::tuple<Tp...>& t)
{ }

template<std::size_t I = 0, typename... Tp>
 typename std::enable_if<I < sizeof...(Tp), void>::type
    print(const std::tuple<Tp...>& t)
{
    std::cout << std::get<I>(t) << std::endl;
    print <i, Tp...> (t);
}

this compiles and works perfectly on GCC, but fails to compile on VC++ (I use visual studio 2013). the error I'm gettting :

Error   4   error C2893: Failed to specialize function template 'std::enable_if<I==1,void>::type print(const std::tuple<_Types1...> &)' 
Error   3   error C2770: invalid explicit template argument(s) for 'std::enable_if<I<1,void>::type print(const std::tuple<_Types1...> &)'

appearntly there is a documented bug on C2770 when using std::enable_if with explicit template-arguments. some developers recommended using const int as pre-argument for the template as

const int i = I+1;
print<i,Tp...>(t);

but this does not work either. there were other solutions also such as use some macros , but they fail too.
does someone have a work around it ? I searched for a solution but found none that actually works.
thanks.

Works for me in VS2015 RC. Release is (hopefully) soon, so toolchain upgrade could be non-standard, but viable option too. (Not even mentioning `constexpr`, faster compilation and linkage etc. in VS2015) — Ivan Aksamentov - Drop, May 04 '15 at 16:16

Jarod42 · Accepted Answer · 2017-08-22T08:50:21.447

You may use one of the following:

using decrementing recursion and (partial) specialization:

namespace detail
{
    template <std::size_t N>
    struct printer
    {
        template <typename TUPLE>
        void operator () (const TUPLE& t) const
        {
            printer<N - 1>{}(t);
            std::cout << std::get<N - 1>(t) << std::endl;
        }
    };

    template <>
    struct printer<0>
    {
        template <typename TUPLE>
        void operator () (const TUPLE& t) const {}
    };

}


template <typename ... Ts>
void print(const std::tuple<Ts...>& t)
{
    detail::printer<sizeof...(Ts)>{}(t);
}

or using index_sequence

#if 1 // Not in C++11
#include <cstdint>

template <std::size_t...> struct index_sequence {};

template <std::size_t N, std::size_t... Is>
struct make_index_sequence : make_index_sequence<N - 1, N - 1, Is...> {};

template <std::size_t... Is>
struct make_index_sequence<0u, Is...> : index_sequence<Is...>{};

#endif

namespace detail
{

    template <std::size_t... Is, typename TUPLE>
    void print(const TUPLE& t, index_sequence<Is...>)
    {
        int dummy[] = {0, ((std::cout << std::get<Is>(t) << std::endl), 0)...};
        (void) dummy; // To remove warning about unused variable.
    }

}

template <typename ... Ts>
void print(const std::tuple<Ts...>& t)
{
    detail::print(t, make_index_sequence<sizeof...(Ts)> {});
}

I didn't tried the second code snippet but the first defintly works. thanks. — David Haim, May 04 '15 at 15:40

score 1 · Answer 2 · answered Jan 11 '16 at 05:54

To retain the structure of the original code given in the question, the following workaround works for me in MSVC 2013:

Define the following helper:

// workaround for msvc `sizeof...(Tp)` not working
template<typename... Tp>
struct sizeof_pack___ { static const std::size_t value = sizeof...(Tp); };

Then replace the two occurrences of sizeof...(Tp) with sizeof_pack___<Tp...>::value.

The final code would look like this:

template<std::size_t I = 0, typename... Tp>
 typename std::enable_if<I == sizeof_pack___<Tp...>::value, void>::type
    print(const std::tuple<Tp...>& t)
{ }

template<std::size_t I = 0, typename... Tp>
 typename std::enable_if<I < sizeof_pack___<Tp...>::value, void>::type
    print(const std::tuple<Tp...>& t)
{
    std::cout << std::get<I>(t) << std::endl;
    print <i, Tp...> (t);
}

Visual studio - workaround error c2770 on std::enable_if

2 Answers2