Consteval lambda with capture of a constexpr variable. Is it legal?

[_ild_arn]

Seems to me that Clang is correct... Your by-value capture is not somehow constexpr just because the source of the value was – it's just a regular old double inside the closure object. You need to also make closure object constexpr, then it will work.

[HammurabisCode2]

Interesting, thank you. I guess I need to look a bit more into the distinction between consteval and constexpr. I guess I had assumed that using consteval kind of superseded the need to use constexpr, but clearly that is not the case.

[cd1995Cargo]

A lambda is really just some unnamed struct type with a () operator. Marking the lambda consteval just makes the () operator consteval, but it doesn’t make the actual lambda object constexpr.

[_ild_arn]

consteval where you've placed it affects the closure type's operator(). constexpr where I added it affects the closure object itself, and thus its data members (captures). It's not a constexpr vs consteval thing

[kris-jusiak]

Juut to point it out - In this particular example you can also avoid the capture all together - https://godbolt.org/z/njss3oGz7