std::index_sequence_for作为默认参数的问题?

Question

下面的c++ 20程序:

#include <utility>
#include <cstddef>
template<typename... Args>
class C {
template<size_t... I>
static void call(
std::index_sequence<I...> = std::index_sequence_for<Args...>{}
) {}
};
int main() {
C<long int>::call();
}

编译失败，错误信息:

test.cc: In static member function ‘static void C<Args>::call(std::index_sequence<I ...>) [with long unsigned int ...I = {}; Args = {long int}; std::index_sequence<I ...> = std::integer_sequence<long unsigned int>]’:
test.cc:11:20: error: could not convert ‘std::index_sequence_for<long int>{}’ from ‘integer_sequence<[...],#‘nontype_argument_pack’ not supported by dump_expr#<expression error>>’ to ‘integer_sequence<[...],#‘nontype_argument_pack’ not supported by dump_expr#<expression error>>’
11 |  C<long int>::call();
|                    ^
|                    |
|                    integer_sequence<[...],#‘nontype_argument_pack’ not supported by dump_expr#<expression error>>
test.cc:11:20: note:   when instantiating default argument for call to ‘static void C<Args>::call(std::index_sequence<I ...>) [with long unsigned int ...I = {}; Args = {long int}; std::index_sequence<I ...> = std::integer_sequence<long unsigned int>]’
test.cc: In function ‘int main()’:
test.cc:11:20: error: could not convert ‘std::index_sequence_for<long int>{}’ from ‘integer_sequence<[...],#‘nontype_argument_pack’ not supported by dump_expr#<expression error>>’ to ‘integer_sequence<[...],#‘nontype_argument_pack’ not supported by dump_expr#<expression error>>’

任何想法?

更新:

我目前最好的解决方法是将默认参数分解为两个函数，如:

template<typename... Args>
class C {
static void call() {
_call(std::index_sequence_for<Args...>{});
}
template<size_t... I>
static void _call(std::index_sequence<I...>) {}
};

这似乎可以解决编译器错误(如果是这样的话)。

更新2:

下面的程序失败的原因与原来的程序相同:

template<typename T> void f(T x = 42) {}
int main() { f(); }

所以这是一个特性而不是一个bug。

Answer 1

一般来说，模板参数演绎+默认函数参数会造成很多麻烦。要简单地修复它，你可以这样做:

#include <utility>
#include <cstddef>
template<typename... Args>
class C {
public:
static void call() {
call_impl(std::index_sequence_for<Args...>{});
}
private:
template<size_t... I>
static void call_impl(std::index_sequence<I...> ) {
}
};
int main() {
C<long int>::call();
}

Answer 2

在c++ 20中，您还可以编写一个模板化的lambda来完成此操作，而无需创建新函数:

//...
static void call() {
return [&]<size_t... Is>(std::index_sequence<Is...>) {
/* your code goes here... */ 
}( std::index_sequence_for<Args...>{} );
}