This is just how constexpr if works. If we check [stmt.if]/2 If the if statement is of the form if constexpr, the value of the condition shall be a contextually converted constant expression of type bool; this form is called a constexpr if statement. If the value of the converted condition is false, the first … Read more
This is just how constexpr if works. If we check [stmt.if]/2 If the if statement is of the form if constexpr, the value of the condition shall be a contextually converted constant expression of type bool; this form is called a constexpr if statement. If the value of the converted condition is false, the first … Read more
Will consteval allow to use static_assert on function arguments? No. Function arguments have never been, and will continue to not be, usable as constant expressions. There is a difference between something being constant evaluated and being usable as a constant-expression. consteval ensures that we’re in a constant evaluation context, but it does not also cause … Read more
Based on example from Jacksonville’18 ISO C++ Report the syntax is correct, but it seems that it is not implemented yet: if (a>b) [[likely]] { 10.6.6 Likelihood attributes [dcl.attr.likelihood] draft
One of the new features in C++20 is Down with typename. In C++17, you had to provide the typename keyword in nearly all†dependent contexts to disambiguate a type from a value. But in C++20, this rule is relaxed a lot. In all contexts where you need to have a type, the typename keyword is … Read more
In order to provide the amortized constant time complexity required by range, filter_view::begin caches the result in *this. This modifies the internal state of *this and thus cannot be done in a const member function.
Here’s another solution, which is more generic (applicable to any expression, without defining a separate template each time). This solution leverages that (1) lambda expressions can be constexpr as of C++17 (2) the type of a captureless lambda is default constructible as of C++20. The idea is, the overload that returns true is selected when … Read more
In C++20 the way that the relational operators work was changed, notably with the introduction of the spaceship <=> operator. In particular, If you only provide operator==, then a != b is rewritten to !(a == b). From [over.match.oper]/3.4: The rewritten candidate set is determined as follows: For the relational ([expr.rel]) operators, the rewritten candidates … Read more
why can’t filter just pass the value to whoever is after it in the pipeline as soon as it finds a match? Because in the iterator model, positioning and accessing are distinct operations. You position an iterator with ++; you access an iterator with *. These are two distinct expressions, which are evaluated at two … Read more
This shouldn’t be valid. The point of allowed lambdas into unevaluated contexts wasn’t to suddenly allow SFINAE on statements. We do have some wording in [temp.deduct]/9 that makes this clear: A lambda-expression appearing in a function type or a template parameter is not considered part of the immediate context for the purposes of template argument … Read more