c++ - 重载可变长度模板函数的递归结束

Question

François Andrieux给了我一个很好的解决方法 this Visual Studio 2017 problem .我试图以他的回答为基础:

template<class T, size_t N>
ostream& vector_insert_impl(ostream& lhs, const char*, const T& rhs)
{
    return lhs << at(rhs, N);
}

template<class T, size_t N, size_t... I>
ostream& vector_insert_impl(ostream& lhs, const char* delim, const T& rhs)
{
    return vector_insert_impl<T, I...>(lhs << at(rhs, N) << delim, delim, rhs);
}

template <typename T, size_t... I>
ostream& vector_insert(ostream& lhs, const char* delim, const T& rhs, index_sequence<I...>) 
{
    return vector_insert_impl<T, I...>(it, delim, rhs);
}

关键区别在于“递归结束”模板化函数实际上将最后一个值插入到 ostream 中。 , 而不是定界符 而不是空操作。但是当我尝试编译它时，我得到了错误:

error C2668: vector_insert_impl: ambiguous call to overloaded function (compiling source file ....\src\STETestbed\Test.cpp)
note: could be std::ostream &vector_insert_impl<T,2,>(std::ostream &,const char *,const T &)
note: or std::ostream &vector_insert_impl<T,2>(std::ostream &,const char *,const T &)

我认为可变长度模板函数被认为是 3^rd 类公民，固定长度模板函数总是首选。这种偏好在这里似乎没有生效。是否有一种解决方法可以强制编译器选择我的“递归结束”函数，从而避免插入定界符？

Answer 1

Is there a workaround which will force the compiler to choose my "end of recursion" function enabling me to avoid inserting the delimiter?

您可以添加“Next”元素

template <typename T, std::size_t N>
std::ostream & vector_insert_impl (std::ostream & lhs, char const *, T const & rhs)
{
    return lhs << at(rhs, N);
}

// ..................................vvvvvvvvvvvvvvvv
template <typename T, std::size_t N, std::size_t Next, std::size_t ... I>
std::ostream & vector_insert_impl (std::ostream & lhs, char const * delim, T const & rhs)
{ // ............................vvvv
    return vector_insert_impl<T, Next, I...>(lhs << at(rhs, N) << delim, delim, rhs);
}

但是，如果你可以使用 C++17，我想 if constexpr 是一个更好的解决方案

template <typename T, std::size_t N, std::size_t ... Is>
std::ostream & vector_insert_impl (std::ostream & lhs, char const * delim, T const & rhs)
{
   if constexpr ( sizeof...(Is) )
      return vector_insert_impl<T, Is...>(lhs << at(rhs, N) << delim, delim, rhs);
   else
      return lhs << at(rhs, N);
}