从默认参数推导参数包

Question

编译器是否可以从函数的默认参数中推导出参数包？特别是，我有以下代码：


template <int ... Is> struct seq {};
template <int ... Is> struct make_seq;
template <int head, int ... tail>
struct make_seq<head, tail...>
{
    using type = typename make_seq<head - 1, head - 1, tail...>::type;
};
template <int ... Is>
struct make_seq<0, Is...>
{
    using type = seq<Is...>;
};
template <int N>
using make_seq_t = typename make_seq<N>::type;

template<int N, int ...Is>
int deduceParamPack(seq<Is...> s = make_seq_t<N>{})
{
    return sizeof...(Is);
}

int main()
{
    return deduceParamPack<5>();
}

并且编译器将参数包推断为空，并尝试将默认参数转换为它。相反，我想实现类似的行为：

int main()
{
    return deduceParamPack<5>(make_seq_t<5>{});
}

推导的参数包是0,1,2,3,4，没有显式传入这个参数。

Answer 1

Is it possible for the compiler to deduce a parameter pack from a function's default aargument?

没有，据我所知

但是......不完全是你问的......但也许你可以找到有用的基于结构部分专业化的以下解决方案

template <std::size_t N, typename = std::make_index_sequence<N*N>>
struct deduceParamPackStruct;

template <std::size_t N, std::size_t ... Is>
struct deduceParamPackStruct<N, std::index_sequence<Is...>>
 {
   static constexpr std::size_t func ()
    { return sizeof...(Is); }
 };

您可以如下使用

static_assert( 25 == deduceParamPackStruct<5>::func() );

从默认参数推导参数包

Deduce parameter pack from default argument

c++

template-meta-programming

variadic-templates

template-argument-deduction

c++17