In C++, is there any way for variatic template to ignore non-arithmetic type or object and return the sum of remain parameters?

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I want to realize a function as the question title described, for example:

cout << SumValue("abc", string("abcd"), 1.3, 1, 10, 2, 100) << endl;

I want that C++ snippet output 114.3, the sum of 1.3, 1, 10, 2 and 100, and ignore "abc" and string("abcd").
I have tried the variatic template function and used the <type_traits> lib as codes below:

template <typename T>
long double SumValue(T first)
{
  if (is_arithmetic<T>::value)
    return first;
  else
    return 0;
}

template <typename T, typename... Args>
long double SumValue(T first, Args... args)
{
  if (is_arithmetic<T>::value)
    return first + SumValue(args...);
  else
    return SumValue(args...);
}

But compiler reports an error:

error: invalid operands of types 'const char*' and 'long double' to binary 'operator+'
     return first + SumValue(args...);
            ~~~~~~^~~~~~~~~~~~~~~~~~~

I can understand the error described, but, I cannot figure out a solution for it.

c++11 templates c++
1 Answers

You can overload SumValue with the help of SFINAE.

template <typename T>
typename std::enable_if<std::is_arithmetic<T>::value, long double>::type
SumValue(T first)
{
  return first;
}
template <typename T>
typename std::enable_if<!std::is_arithmetic<T>::value, long double>::type
SumValue(T first)
{
  return 0;
}

template <typename T, typename... Args>
long double SumValue(T first, Args... args)
{
  return SumValue(first) + SumValue(args...);
}

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BTW: Since C++17 you can use Constexpr If.

template <typename T>
long double SumValue(T first)
{
  if constexpr (std::is_arithmetic<T>::value)
    return first;
  else
    return 0;
}

template <typename T, typename... Args>
long double SumValue(T first, Args... args)
{
  return SumValue(first) + SumValue(args...);
}

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