c++ - Workaround for lack of expression SFINAE -

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i'm trying call function each value in std::tuple, of course there no way iterate tuple , i've resorted using template techniques discussed in iterate on tuple

however, i'm using visual studio 2013 , not support expression sfinae, code won't work. i've tried partially specialize templates based on constant numbers (e.g 5, 4, 3, 2, 1, 0), haven't had success. i'm no template expert, , hoping me out. expression sfinae code below.

#include <iostream> #include <tuple>  using namespace std;  struct argpush {     void push(bool x) {}     void push(int x) {}     void push(double x) {}     void push(const char* x) {}     void push(const std::string& x) {}      template<std::size_t = 0, typename... tp>     inline typename std::enable_if<i == sizeof...(tp), void>::type         push_tuple(const std::tuple<tp...>& t)     { }      template<std::size_t = 0, typename... tp>     inline typename std::enable_if<i < sizeof...(tp), void>::type         push_tuple(const std::tuple<tp...>& t)     {         push(std::get<i>(t));         push_tuple<i + 1, tp...>(t);     } };  int main() {     argpush().push_tuple(std::make_tuple(1,2,3,4));     argpush().push_tuple(std::make_tuple("hello", "msvc makes me sad", 4, true));     return 0; }

msvc doesn't equality comparisons being made within enable_if. move these out there helper template. code compiles on vs2013.

template<std::size_t i, typename... tp> struct comparator {     static const bool value = (i < sizeof...(tp)); };  template<std::size_t = 0, typename... tp> inline typename std::enable_if<!comparator<i, tp...>::value>::type     push_tuple(const std::tuple<tp...>& t) { }  template<std::size_t = 0, typename... tp> inline typename std::enable_if<comparator<i, tp...>::value>::type     push_tuple(const std::tuple<tp...>& t) {     push(std::get<i>(t));     push_tuple<i + 1, tp...>(t); }

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