On Jan 30, 2022, at 7:51 AM, Jonathan O'Connor via SG7 <sg7@lists.isocpp.org> 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:

0. 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@lists.isocpp.org> wrote:

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