关于规范中的§8/ 5，我需要一些帮助

Question

§8/5:

The optional attribute-specifier-seq in a trailing-return-type appertains to the indicated return type. The type-id in a trailing-return-type includes the longest possible sequence of abstract-declarator s. [ Note: This resolves the ambiguous binding of array and function declarators. [ Example:

 auto f()->int(*)[4]; // function returning a pointer to array[4] of int // not function returning array[4] of pointer to int 

—end example ] —end note ]

The " type-id in a trailing-return-type " doesn't make sense to me, simply because a trailing-return-type doesn't contain a type-id according to the grammar.

I also don't understand the "ambiguous binding" of array and function declaration. As far as I can understand

auto f() -> int*[4]; // function returning an array of 4 pointers to int
auto f() -> int(*)[4]; // function returning a pointer to an array of 4 ints  

Answer 1

int *f();

Declares a function of () returning pointer to int .

int *f()[4];

Declares a function of () returning array of 4 pointers to int . Note that this is ill-formed.

int (*f())[4];

Declares a function of () returning pointer to array of 4 int .

Now, in

  auto f() -> int(*)[4]
//     ^^^^^^^^^^^^^---

What the rule resolves is whether [4] is part of the trailing-return-type , and hence part of the the function declarator. If [4] is part of the trailing-return-type , then the above declaration declares a function of () returning pointer to array of 4 int .

If not, then [4] would form an array declarator that is not part of the function declarator, and the parse would be governed by [dcl.array]/p1:

In a declaration TD where D has the form

 D1 [ constant-expression_opt ] attribute-specifier-seq_opt 

and the type of the identifier in the declaration T D1 is “ derived-declarator-type-list T ” [..., if] the value of the constant expression is N , [...] the type of the identifier of D is “ derived-declarator-type-list array of N T ”.

and since auto f()-> int (*) declares f as "function of () returning pointer to int ", substitution tells us that this would declare a function returning an array of 4 pointers to int , just like int *f()[4]; .