不应该std::invoke成为 constexpr 尤其是在 constexpr lambdas in C++17 之后？
是否有任何障碍可以阻止这种情况发生？
最佳答案
更新: P1065将使它成为 constexpr。
由于历史原因保留原帖:
来自 the proposal :
Although there is possibility to implement standard conforming invoke function template as a constexpr function, the proposed wording does not require such implementation. The main reason is to left it consistent with existing standard function objects, that could have such definition, like std::mem_fn, std::reference_wrapper and operator wrappers. Furthermore imposing such requirement will block the implementation of invoke that refers to std::mem_fn.
This proposal assumes that constexpr addition to the header would be applied consistently by a separate proposal.
Both constexpr and standard library based implementation are presented in Implementability section of the proposal.
相关 CWG 问题 #1581:When are constexpr member functions defined? .
关于c++ - 为什么 `std::invoke` 不是 constexpr？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/40222989/