> On Jan 30, 2022, at 7:51 AM, Jonathan O'Connor via SG7 <sg7_at_[hidden]> wrote:
>
> Fellow elevator users,
>
> Page 5:
>
> In the discussion of `^f(x)` it is not clear if `f(x)` is evaluated.

The sentence after the bullets describing what operands can be provided to the lifting operator says:
In the case where the name_or_postfix_expr is an expression, it is unevaluated but potentially constant evaluated.



> From later discussion,
> it would seem that it is evaluated if f(x) is constexpr. Presumably `^printf("hello")` is not evaluated.
>
> The final paragraph does not mention applying the lift operator to a lambda.

I’m not sure which final paragraph you’re referring to?

In C++ lambda-expressions are expressions, and a lambda-expression is a postfix-expression (via primary-expression). So, yes, you can reflect a lambda-expression.

(We don’t list every possible expansion of the grammar for purposes of conciseness.)


> Is
> that possibility missing, or is it included as a class?

I’m not sure what you mean by “is it included as a class?”. A lambda-expression is not a class, but it has class type. P1240 allows querying the type of an expression in general and so that includes lambda-expressions. So you can for example write:

 typename[: type_of(^[]{}) :] x;

and that will produce a variable x of a closure type corresponding to that of the reflected lambda-expression.


>
> Page 9: final code block:
>
> There should be a comment on `^printf("Hello, ")` stating the printf function is not invoked
> before the lift operator.
>

It seems a bit redundant, but sure, I can add that. (Done in my version now.)

 Daveed

>
> Regards,
> Jonathan
>
> ‐‐‐‐‐‐‐ Original Message ‐‐‐‐‐‐‐
>
> On Saturday, January 15th, 2022 at 17:41, Daveed Vandevoorde via SG7 <sg7_at_[hidden]> wrote:
>
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