(int i : std::views::iota(1, Bound{10})) std::cout << i << ' '; std::cout << '\n'; for (int i : std::views::iota(1) | std::views::take(9)) std::cout << i << ' '; std::cout << '\n'; std::ranges::for_each(std::views::iota(1, 10), [](int i) ...
std::ranges::iota_view::base_ W value_=W();/* exposition-only */ the current value std::ranges::iota_view::pred_ Bound bound_=Bound();/* exposition-only */ the bound (defaults tostd::unreachable_sentinel_t) Member functions
std::ranges::iota(v,0); for(inti=v.size-1;i>=0;--i){ std::cout<<v[i]<<''; } 乍看之下,似乎并无问题,但实际上却存在安全隐患,若是 v.size 的结果大于 std::numeric_limits<int>::max,将产生 UB。 倘若你使用了类型推导,问题会更加明显。 for(autoi=v.size-1;i>=0;--i){ std:...
I tried to create a std::ranges::iota_view with a value and a bound: #include <ranges> #include <vector> int main() { std::vector v{1,2,3}; auto indices = std::ranges::views::iota(0, v.size()); } link to program on godbolt But clang 11.0.0 fails to compile w...
operator-(iota_view::iterator,iota_view::sentinel) constexpr/*iterator*/begin()const; (since C++20) Returns aniteratorinitialized withvalue_. Parameters (none) Return value Aniterator Example Run this code #include <iostream>#include <ranges>intmain(){autoiota{std::views::iota(2,6)};autoite...
std::sort(v.begin(),v.end());std::ranges::sort(v);std::transform(m.begin(),m.end(),...
问题是,我有一些通用代码,我想向它发送一个范围,然后在该范围上创建一个跨度。我试过发送一个vector,没问题。结果iota失败。template <typename TRange> requires std::ranges::random_access_range<TRange> void Foo(TRange const & r) { // The algorithm starts with a full span and then partitions auto...
Applyingstd::ranges::upper_boundover an iota view with along longvalue results in compilation failure when importing the standard library via thestdmodule instead of using the standard library headers: importstd;voidtest(){automax = std::ranges::upper_bound(std::views::iota(0LL,...
Usingstd::ranges::views::iotaused to initialise an explicit specialisation of a variable template causes an internal compiler error. For example: #include <ranges> template<typename T> auto foo = 1; template<> auto foo<int> = std::ranges::views::iota(1, 10); int main() {}...
2. 由于随着范围库引入了不可变范围，例如std::views::iota()创建的范围是不可变范围，它不适合要求它...