标准中的哪个地方表示声明`auto f（）（） - > int;`是不允许的？

Question

I know I'm being pedantic here, but I'm just trying to understand the C++ grammar production.

I will start with a simple-declaration .

simple-declaration :
decl-specifier-seq _opt init-declarator-list _opt

decl-specifier-seq :
decl-specifier

decl-specifier :
type-specifier

type-specifier :
trailing-type-specifier

trailing-type-specifier :
simple-type-specifier

simple-type-specifier :
char
int
...
auto

---

Now, I'll look into the definition of a init-declarator-list .

init-declarator-list :
init-declarator

init-declarator :
declarator

declarator :
noptr-declarator parameters and qualifiers trailing return type (*2)

noptr-declarator :
declarator-id attribute-specifier-seq _opt
noptr-declarator parameters and qualifiers (*1)

parameters and qualifiers :
( parameter-declaration-clause ) cv-qualifiers _opt ... (all optionals)

trailing-return-type :
- > trailing-type-specifier-seq

trailing-type-specifier-seq :
trailing-type-specifier See the definition of a traling-type-specifier above.

---

Replacing noptr-declarator by a declarator-id in (*1) , using the previous definition of noptr-declaration , we arrive at the following definition of nonptr-declarator :

noptr-declaration :
declarator-id parameters and qualifiers

Now replacing noptr-declarator , parameters and qualifiers and trailing-return-type in (*2) by each definition given above, we obtain the following for declarator :

declarator :
declarator-id ( parameter-declaration-clause ) ( parameter-declaration-clause ) - >
simple-type-specifier

---

With this last result we could say that the grammar allows the following declaration of a function f :

auto f()() -> int;  

which of course is not valid. But I couldn't find anything in the Standard saying, directly or indirectly, that this construction is ill-formed.

The error messages from GCC ( f declared as function returning a function) and clang ( auto return without trailing return type; deduced return types are a C++1y extensions) didn't help in this regard either.

Answer 1

[dcl.fct] /8:

[...] Functions shall not have a return type of type array or function, although they may have a return type of type pointer or reference to such things. [...]

Answer 2

In fact, the gcc error message is quite helpful: The grammar does in fact allow auto f()() -> int , but this syntactically correct declaration is semantically invalid. It would describe a function returning a function returning int . See the answer of ecatmur for the standard quote disallowing that.

To understand the parse, work like this. f is what you want to declare. Parse to the right, you'll find an empty argument list, so f() is a valid expression, and the type of that expression gets declared. Parse again to the right, you'll find an argument list again , so the return type of f is a function taking zero arguments. Parse again to the right, and you will hit the end (the -> arrow), so parse to the left to find the result type, which is auto, and gets replaced by the type past the arrow.