constexpr

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 Supposedly std::abs is not constexpr in the standard (even in C++20). But in practice I found out that I can compile it as constexpr under the very peculiar condition that the function is templated. See this completely working example: template<class T> constexpr T f(const T input) { return std::abs(input); } int main() { int i = -1; int a = f(i); return 0; } The code: Compiles fine with GCC, with and without the template. It doesn't work in Clang. And in Visual Studio it compiles with the

问题 This program: constexpr void f() { x: ; } is compiled by gcc, but clang says: error: statement not allowed in constexpr function So is this code valid? 回答1: As pointed out in a comment, clang is correct, and the code is ill-formed. According to the current working draft (which includes C++20), dcl.constexpr#3 says: The definition of a constexpr function shall satisfy the following requirements: ... its function-body shall not enclose ... an identifier label, ... ... 来源： https://stackoverflow

问题 I'm implementing a constexpr int foo(); function. In the body of foo() , I want to do something different (and possibly return something different) at compile time and something different at run-time. With C++20, I can use std::is_constant_evaluated: constexpr int foo() { return std::is_constant_evaluated() ? 123 : 456 }; but what if I'm using C++17 (or earlier) - what can I do with the same effect? Note: Compiler-specific solutions are acceptable (though less desirable). 回答1: Note: Compiler

问题 I'm implementing a constexpr int foo(); function. In the body of foo() , I want to do something different (and possibly return something different) at compile time and something different at run-time. With C++20, I can use std::is_constant_evaluated: constexpr int foo() { return std::is_constant_evaluated() ? 123 : 456 }; but what if I'm using C++17 (or earlier) - what can I do with the same effect? Note: Compiler-specific solutions are acceptable (though less desirable). 回答1: Note: Compiler

问题 I'm implementing a constexpr int foo(); function. In the body of foo() , I want to do something different (and possibly return something different) at compile time and something different at run-time. With C++20, I can use std::is_constant_evaluated: constexpr int foo() { return std::is_constant_evaluated() ? 123 : 456 }; but what if I'm using C++17 (or earlier) - what can I do with the same effect? Note: Compiler-specific solutions are acceptable (though less desirable). 回答1: Note: Compiler

来源： https://stackoverflow.com/questions/53419133/constexpr-operator-new

来源： https://stackoverflow.com/questions/27498592/c11-get-string-length-in-compile-time-by-constexpr

来源： https://stackoverflow.com/questions/27498592/c11-get-string-length-in-compile-time-by-constexpr