c++20

问题 I was looking through 'Atomic operations library' and came across a new c++20 feature of atomic 'wait' and 'notify_ ' methods. I am curious on what the differences are in regards to std::condition_variable's 'wait' and 'notify_ ' methods. 回答1: std:atomic wait , notify_all and notify_one methods are similar to methods of conditional variables. They allow the implementation of the logic that previously required conditional variable by using much more efficient and lightweight atomic variables.

问题 I was looking through 'Atomic operations library' and came across a new c++20 feature of atomic 'wait' and 'notify_ ' methods. I am curious on what the differences are in regards to std::condition_variable's 'wait' and 'notify_ ' methods. 回答1: std:atomic wait , notify_all and notify_one methods are similar to methods of conditional variables. They allow the implementation of the logic that previously required conditional variable by using much more efficient and lightweight atomic variables.

问题 I was looking through 'Atomic operations library' and came across a new c++20 feature of atomic 'wait' and 'notify_ ' methods. I am curious on what the differences are in regards to std::condition_variable's 'wait' and 'notify_ ' methods. 回答1: std:atomic wait , notify_all and notify_one methods are similar to methods of conditional variables. They allow the implementation of the logic that previously required conditional variable by using much more efficient and lightweight atomic variables.

问题 I was under the impression that the following should become valid code under the new C++20 standard: struct Foo { int a, b; }; template<Foo> struct Bar {}; Bar<{.a=1, .b=2}> bar; Yet, gcc 10.2.0 , with -std=c++20 set complains: could not convert ‘{1, 2}’ from ‘<brace-enclosed initializer list>’ to ‘Foo’ and Clang cannot compile this snippet either. Can someone point out why it is not well formed? 回答1: This template-argument {.a=1, .b=2} is not allowed according to the grammar for a template

问题 I was under the impression that the following should become valid code under the new C++20 standard: struct Foo { int a, b; }; template<Foo> struct Bar {}; Bar<{.a=1, .b=2}> bar; Yet, gcc 10.2.0 , with -std=c++20 set complains: could not convert ‘{1, 2}’ from ‘<brace-enclosed initializer list>’ to ‘Foo’ and Clang cannot compile this snippet either. Can someone point out why it is not well formed? 回答1: This template-argument {.a=1, .b=2} is not allowed according to the grammar for a template

问题 As constexpr std::string and constexpr std::vector have been accepted into C++20, how will these be used? The linked papers are very short on details. Do we need to specify special constexpr allocators, making compile-time strings/vectors incompatible with their normal equivalents? 回答1: Those two papers depend heavily on P0784, which discusses how allocations at compile-time will work. Incomplete answer: Only std::allocator will work. All allocations are tracked, and must be deallocated

问题 As constexpr std::string and constexpr std::vector have been accepted into C++20, how will these be used? The linked papers are very short on details. Do we need to specify special constexpr allocators, making compile-time strings/vectors incompatible with their normal equivalents? 回答1: Those two papers depend heavily on P0784, which discusses how allocations at compile-time will work. Incomplete answer: Only std::allocator will work. All allocations are tracked, and must be deallocated

问题 As constexpr std::string and constexpr std::vector have been accepted into C++20, how will these be used? The linked papers are very short on details. Do we need to specify special constexpr allocators, making compile-time strings/vectors incompatible with their normal equivalents? 回答1: Those two papers depend heavily on P0784, which discusses how allocations at compile-time will work. Incomplete answer: Only std::allocator will work. All allocations are tracked, and must be deallocated

问题 As constexpr std::string and constexpr std::vector have been accepted into C++20, how will these be used? The linked papers are very short on details. Do we need to specify special constexpr allocators, making compile-time strings/vectors incompatible with their normal equivalents? 回答1: Those two papers depend heavily on P0784, which discusses how allocations at compile-time will work. Incomplete answer: Only std::allocator will work. All allocations are tracked, and must be deallocated

问题 I wanted to create a deep_flatten function template that would produce a range of elements that are deeply join ed. For example, if we take into account only nested std::vector s, I can have: template <typename T> struct is_vector : public std::false_type { }; template <typename T, typename A> struct is_vector<std::vector<T, A>> : public std::true_type { }; template <typename T> auto deepFlatten(const std::vector<std::vector<T>>& vec) { using namespace std::ranges; if constexpr (is_vector<T>: