當試用方便的方式來訪問元組作爲容器時,我寫了一個測試程序。哪個編譯器(如果有的話)在參數包擴展中有缺陷?
上鐺(3.9.1,和蘋果鐺)它編譯如預期,產生預期的輸出:
1.1
foo
2
上的gcc(5.4,6.3),它不能編譯:
<source>: In lambda function:
<source>:14:61: error: parameter packs not expanded with '...':
+[](F& f, Tuple& tuple) { f(std::get<Is>(tuple)); }...
^
<source>:14:61: note: 'Is'
<source>: In function 'decltype(auto) notstd::make_callers_impl(std::index_sequence<Is ...>)':
<source>:14:64: error: expansion pattern '+<lambda>' contains no argument packs
+[](F& f, Tuple& tuple) { f(std::get<Is>(tuple)); }...
^~~
Compiler exited with result code 1
問題:誰是對的?它可以修復嗎?
計劃:
#include <iostream>
#include <array>
#include <tuple>
namespace notstd {
template<class F, class Tuple, std::size_t...Is>
auto make_callers_impl(std::index_sequence<Is...>) -> decltype(auto)
{
static std::array<void (*) (F&, Tuple&), sizeof...(Is)> x =
{
+[](F& f, Tuple& tuple) { f(std::get<Is>(tuple)); }...
};
return x;
};
template<class F, class Tuple>
auto make_callers() -> decltype(auto)
{
return make_callers_impl<F, Tuple>(std::make_index_sequence<std::tuple_size<std::decay_t<Tuple>>::value>());
};
template<class Tuple, std::size_t N = std::tuple_size<std::decay_t<Tuple>>::value >
struct tuple_iterator {
static constexpr auto size = N;
constexpr tuple_iterator(Tuple& tuple, std::size_t i = 0) : tuple(tuple), i(i) {}
template<class F>
void with(F&& f) const {
static const auto& callers = make_callers<F, Tuple>();
callers[i](f, tuple);
}
constexpr bool operator!=(tuple_iterator const& r) const {
return i != r.i;
}
constexpr auto operator++() -> tuple_iterator& {
++i;
return *this;
}
Tuple& tuple;
std::size_t i;
};
template<class Tuple>
auto begin(Tuple&& tuple)
{
return tuple_iterator<Tuple>(std::forward<Tuple>(tuple));
}
template<class Tuple>
auto end(Tuple&& tuple)
{
using tuple_type = std::decay_t<Tuple>;
static constexpr auto size = std::tuple_size<tuple_type>::value;
return tuple_iterator<Tuple>(std::forward<Tuple>(tuple), size);
}
}
template<class T> void emit(const T&);
int main() {
auto a = std::make_tuple(1.1, "foo", 2);
auto i = notstd::begin(a);
while(i != notstd::end(a))
{
i.with([](auto&& val) { std::cout << val << std::endl; });
++i;
}
}
我建議你在這個問題上添加標籤'language-lawyer',因爲它是關於編譯器符合標準 –
@GuillaumeRacicot這樣做的。謝謝。 –
嗯,我猜想,鏗鏘是對的,因爲代碼編譯和按預期工作,因爲海灣合作委員會正在拋出編譯器錯誤,它不能在當前版本中執行或修復。 – chbchb55