How to declare a lambda's operator() as noreturn?

Question

How can the operator() of a lambda be declared as noreturn ?

Ideone accepts the following code:

#include   
         


        

Answer 1

Clang is correct. An attribute can appertain to a function being declared, or to its type; the two are different. [[noreturn]] must appertain to the function itself. The difference can be seen in

// [[noreturn]] appertains to the entity that's being declared
void f [[noreturn]] ();    // §8.3 [dcl.meaning]/p1:
                           // The optional attribute-specifier-seq following a
                           // declarator-id appertains to the entity that is declared."
[[noreturn]] void h ();    // §7 [dcl.dcl]/p2:
                           // "The attribute-specifier-seq in a simple-declaration 
                           // appertains to each of the entities declared by
                           // the declarators of the init-declarator-list."

// ill-formed - [[noreturn]] appertains to the type (§8.3.5 [dcl.fct]/p1: 
// "The optional attribute-specifier-seq appertains to the function type.")
void g () [[noreturn]] {}

Indeed if you compile this in g++ it tells you that

warning: attribute ignored [-Wattributes]
 void g () [[noreturn]] {}
                      ^
note: an attribute that appertains to a type-specifier is ignored

Note that it doesn't emit a warning that g() actually does return.

Since an "attribute-specifier-seq in the lambda-declarator appertains to the type of the corresponding function call operator or operator template" (§5.1.2 [expr.prim.lambda]/p5) rather than to that operator/operator template itself, you can't use [[noreturn]] there. More generally, the language provides no way for you to apply an attribute to the operator () of a lambda itself.