But I can say right now the “struct construct {};” is meant to work with all types of constructors, including those with and without a constexpr constructor. That’s why it needs to become a qualifier so that operator () can be overloaded properly.

The goal is to keep the constexpr qualifier as long as possible given the case the arbitrary class uses a constexpr constructor, this way the code will be more efficient...

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**Phil Bouchard**

Founder

C.: (819) 328-4743

Founder

C.: (819) 328-4743

On Thu, Oct 3, 2019 at 12:26 AM Phil Bouchard <phil@fornux.com> wrote:Verdict?

Phil, you should pause and try to come up with a motivating example for the feature you claim to want.Once you have a motivating example, the next step would be to look at what would be the best way to solve it. Maybe there's even a better way than what you originally proposed as a feature!That is, start with a problem, and then propose a solution for the problem. If (by thinking) you end up realizing that you don't have a problem after all, that's actually agoodthing.You do need toabout your examples. Here's your latest one:slow down and thinktemplate <typename T>

struct construct

{

T operator () (node_proxy &, T && t) { return T(t); } // will lose constexpr

};And here's the perfectly valid C++11 code that solves your stated problem:

template <typename T>

struct construct

{

constexprT operator () (node_proxy &, T && t) { return T(t); } // no longer loses constexpr

};See also: http://sscce.org–Arthur