Declarations in C++

Question

From what I have understood, declarations/initializations in C++ are statements with \'base type\' followed by a comma separated list of declarators.

Consider the follow

Answer 1

This is specified in [dcl.dcl] and [dcl.decl] as part of the simple-declaration* and boils down to differences between the branches in ptr-declarator:

declaration-seq:
    declaration

declaration:
    block-declaration

block-declaration:
    simple-declaration

simple-declaration:
    decl-specifier-seqopt init-declarator-listopt ;
----

decl-specifier-seq:
    decl-specifier decl-specifier-seq    

decl-specifier:    
    type-specifier                               ← mentioned in your error

type-specifier:
    trailing-type-specifier

trailing-type-specifier:
    simple-type-specifier
    cv-qualifier
----

init-declarator-list:
   init-declarator
   init-declarator-list , init-declarator

init-declarator:
   declarator initializeropt

declarator:
    ptr-declarator

ptr-declarator:                                 ← here is the "switch"
    noptr-declarator
    ptr-operator ptr-declarator

ptr-operator:                                   ← allows const
    *  cv-qualifier-seq opt

cv-qualifier:
    const
    volatile

noptr-declarator:                               ← does not allow const
    declarator-id

declarator-id:
    id-expression

The important fork in the rules is in ptr-declarator:

ptr-declarator:
    noptr-declarator
    ptr-operator ptr-declarator

Essentially, noptr-declarator in your context is an id-expression only. It may not contain any cv-qualifier, but qualified or unqualified ids. However, a ptr-operator may contain a cv-qualifier.

This indicates that your first statement is perfectly valid, since your second init-declarator

 *const p = &i;

is a ptr-declarator of form ptr-operator ptr-declarator with ptr-operator being * const in this case and ptr-declarator being a unqualified identifier.

Your second statement isn't legal because it is not a valid ptr-operator:

 const c = 2

A ptr-operator must start with *, &, && or a nested name specifier followed by *. Since const c does not start with either of those tokens, we consider const c as noptr-declarator, which does not allow const here.

Also, why the behaviour differs among 3rd and 4th statements?

Because int is the type-specifier, and the * is part of the init-declarator,

*const p1

declares a constant pointer.

However, in int const, we have a decl-specifier-seq of two decl-specifier, int (a simple-type-specifier) and const (a cv-qualifier), see trailing-type-specifier. Therefore both form one declaration specifier.

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_{* Note: I've omitted all alternatives which cannot be applied here and simplified some rules. Refer to section 7 "Declarations" and section 8 "Declarators" of C++11 (n3337) for more information.}