c++14

问题 I got to know the reason that future returned from std::async has some special shared state through which wait on returned future happened in the destructor of future. But when we use std::pakaged_task , its future does not exhibit the same behavior. To complete a packaged task, you have to explicitly call get() on future object from packaged_task . Now my questions are: What could be the internal implementation of future (thinking std::async vs std::packaged_task )? Why the same behavior was

问题 I get compiler error by Clang 3.8 and GCC 5.3 if I want to declare my default -ed default constructors as constexpr . According to this stackoverflow question it just should work fine: struct A { constexpr A() = default; int x; }; however: Error: defaulted definition of default constructor is not constexpr Have you got any clue what is actually going on? 回答1: As it stands, x remains uninitialized, so the object can not be constructed at compile time. You need to initialize x: struct A {

问题 I get compiler error by Clang 3.8 and GCC 5.3 if I want to declare my default -ed default constructors as constexpr . According to this stackoverflow question it just should work fine: struct A { constexpr A() = default; int x; }; however: Error: defaulted definition of default constructor is not constexpr Have you got any clue what is actually going on? 回答1: As it stands, x remains uninitialized, so the object can not be constructed at compile time. You need to initialize x: struct A {

问题 c++14 introduced generic lambdas that made it possible to write following: auto func = [](auto a, auto b){ return a + b; }; auto Foo = func(2, 5); auto Bar = func("hello", "world"); It is very clear that this generic lambda func works just like a templated function func would work. Why did the C++ committee decide to add template syntax for generic lamda? 回答1: C++14 generic lambdas are a very cool way to generate a functor with an operator () that looks like this: template <class T, class U>

问题 c++14 introduced generic lambdas that made it possible to write following: auto func = [](auto a, auto b){ return a + b; }; auto Foo = func(2, 5); auto Bar = func("hello", "world"); It is very clear that this generic lambda func works just like a templated function func would work. Why did the C++ committee decide to add template syntax for generic lamda? 回答1: C++14 generic lambdas are a very cool way to generate a functor with an operator () that looks like this: template <class T, class U>

来源： https://stackoverflow.com/questions/35285913/when-should-we-use-parenthesis-vs-initializer-syntax-to-initialize-obje

来源： https://stackoverflow.com/questions/41519450/what-is-a-generalized-lambda-capture-and-why-was-it-created

来源： https://stackoverflow.com/questions/41519450/what-is-a-generalized-lambda-capture-and-why-was-it-created

来源： https://stackoverflow.com/questions/54655117/adding-googletest-to-existing-cmake-project

来源： https://stackoverflow.com/questions/27551167/what-is-the-endianness-of-binary-literals-in-c14