1

可能重复：
使用带有可变参数模板函数的 decltype 的尾随返回类型

我收到此编译器错误：

g++ -std=gnu++0x -I. -O3 -Wall sum.cpp
sum.cpp:7:41: sorry, unimplemented: cannot expand ‘Remaining ...’ into a fixed-length argument list
sum.cpp: In function ‘int main(int, const char**)’:
sum.cpp:29:23: error: no matching function for call to ‘sum(int, int, int)’
sum.cpp:29:23: note: candidate is:
sum.cpp:20:6: note: template<class FirstArg, class ... RemainingArgs> decltype (Sum<FirstArg, RemainingArgs ...>::sum(first, sum::args ...)) sum(const FirstArg&, const RemainingArgs& ...)

对于这个sum.cpp：

#include <iostream>
#include <type_traits>

template <typename First, typename... Remaining>
struct Sum {
   static auto sum(const First &arg, const Remaining &... args)
      -> decltype(arg + Sum<Remaining...>::sum(args...))
   {
      return arg + sum(args...);
   }
};

template <typename First>
struct Sum<First>
{
   static First sum(const First &arg) { return arg; }
};

template <typename FirstArg, typename... RemainingArgs>
auto sum(const FirstArg &first, const RemainingArgs &... args)
   -> decltype(Sum<FirstArg, RemainingArgs...>::sum(first, args...))
{
   return Sum<FirstArg, RemainingArgs...>::sum(first, args...);
}

int main(int argc, const char *argv[])
{
   using ::std::cout;
   cout << sum(1, 2, 3) << '\n';

   return 0;
}

我尝试了多种声明此功能的方法。这只是 gcc 4.6.1 的问题吗？

编辑：这是我最终选择的，因为我不太信任::std::common_type模板。它仍然有一个问题，它会+从右到左而不是从左到右绑定。这可能会导致非交换+运算符出现问题。不过，这并不难解决：

#include <iostream>
#include <type_traits>
#include <utility>

namespace {
template<class T> typename ::std::add_rvalue_reference<T>::type val();

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

template<class T, class... P> struct sum_type;
template<class T> struct sum_type<T> : id< T > {};
template<class T, class U, class... P> struct sum_type<T,U,P...>
: sum_type< decltype( val<const T&>() + val<const U&>() ), P... > {};
}

template <typename T>
T sum(const T &&arg)
{
   return ::std::forward<const T>(arg);
}

template <typename FirstArg, typename SecondArg, typename... RemainingArgs>
auto sum(const FirstArg &&first, const SecondArg &&second,
         const RemainingArgs &&... args)
   -> typename sum_type<FirstArg, SecondArg, RemainingArgs...>::type
{
   using ::std::forward;

   return forward<const FirstArg>(first) + \
      sum(forward<const SecondArg>(second),
          forward<const RemainingArgs>(args)...);
}

int main(int argc, const char *argv[])
{
   using ::std::cout;
   cout << sum(1, 2, 3.2) << '\n';

   return 0;
}
4

1 回答 1

6

递归调用的第一个内部sum函数具有错误的签名。

对于运行时可变参数函数，试试这个：

#include <type_traits>
#include <utility>

template <typename T> T sum(T && x) { return std::forward<T>(x); }

template <typename T, typename ...Args>
typename std::common_type<T, Args...>::type sum(T && x, Args &&... args)
{
  return std::forward<T>(x) + sum(std::forward<Args>(args)...);
}


int main()
{
  return sum(1,2,3,4);
}
于 2011-11-13T23:04:25.680 回答