从默认参数推导出参数包
[英]Deduce parameter pack from default argument
问题描述
Is it possible for the compiler to deduce a parameter pack from a function's default aargument?
编译器是否可以从函数的默认参数中推断出参数包？ Particularly, I have the following code: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>();
}
And the compiler deduces the parameter pack as empty, and tries to cast the default argument to it.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。 Instead, I would like to achieve a similar behaviour to:相反，我想实现类似的行为：
int main()
{
return deduceParamPack<5>(make_seq_t<5>{});
}
where the deduced parameter pack is 0,1,2,3,4 , without explicitly passing in this argument.其中推导的参数包是0,1,2,3,4 ，没有显式传入这个参数。
1 个解决方案
解决方案1
1 已采纳 2019-10-20 16:38:40
Is it possible for the compiler to deduce a parameter pack from a function's default aargument?
No, as far I know.不，据我所知。
But... not exactly what you asked... but maybe you can find useful the following solution based in partial specialization of structs但是......不完全是你问的......但也许你可以找到有用的以下基于结构部分专业化的解决方案
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); }
};
You can use it as follows您可以按如下方式使用它
static_assert( 25 == deduceParamPackStruct<5>::func() );
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