On 26/01/2025 15.11, Russell Shaw via Std-Discussion wrote:
> On 27/1/25 00:50, Jens Maurer wrote:
>>
>> On 26/01/2025 14.41, Russell Shaw via Std-Discussion wrote:
>>> The example says:
>>> auto b = &f<int>; // error: f<int> is an immediate function
>>>
>>> But the instantiated function is:
>>>
>>> constexpr int f(int t) {
>>> return t + id(t);
>>> }
>>>
>>> The function parameter scope is '(int t)' (or does it encompass the body too ?)
>>
>> It does encompass the function body, otherwise "t" wouldn't be
>> visible in the body. See [basic.scope.param] p1.
>>
>>> Where is the immediate-escalating expression E such that E ’s innermost
>>> enclosing non-block scope is F ’s function parameter scope ?
>>
>> "id(t)" is the expression.
>
> Ok.
> [basic.scope.param] adds no useful information.

How so? It says

"If P is associated with a declarator and is preceded by a (possibly-parenthesized) noptr-declarator of the
form declarator-id attribute-specifier-seq opt , its scope extends to the end of the nearest enclosing [...]
function-definition, but [...].

Note the grammar non-terminal "function-definition" there, which covers the
function body per the grammar in [dcl.fct.def.general].

> [basic.scope.block] says that the function parameter scope is the parent scope
> of the function body (if my understanding that the function body is classed as a
> compound statement and thus a block scope is correct ?)

The function body might be a function-try-block, which is more than
a compound-statement. See [dcl.fct.def.general].
 Jens