Same const type qualifier

Question

Having two const's for a type issues a warning / error. However if the type has been defined with typedef, the compiler accepts it (both Visual Studio 2013 and the online compiler C++ shell).

#include <iostream>  typedef const int value_type;  int main() {     const value_type  n = 0;   //ok      const const int   n2 = 0;  //error C4114      return 0; } 

Does anyone have an idea as to why? Is it that one is const (const int), which is different from const const int?

Answer 1

It's explicitly allowed in the typedef case, and disallowed in the declaration itself:

[dcl.type/1]

The type-specifiers are:

type-specifier : ... cv-qualifier
defining-type-specifier : type-specifier

[dcl.spec/1 and 2]

The specifiers that can be used in a declaration are:

decl-specifier : ... defining-type-specifier ...

Each decl-specifier shall appear at most once in a complete decl-specifier-seq, except that long may appear twice.

[dcl.type.cv/1]

There are two cv-qualifiers, const and volatile. Each cv-qualifier shall appear at most once in a cv-qualifier-seq. If a cv-qualifier appears in a decl-specifier-seq, the init-declarator-list or member-declarator-list of the declaration shall not be empty. [ Note: [basic.type.qualifier] and [dcl.fct] describe how cv-qualifiers affect object and function types.  — end note ] Redundant cv-qualifications are ignored. [ Note: For example, these could be introduced by typedefs. — end note ]

Besides type aliases, a template parameter is another case where the qualifier could be redundant. The rationale for allowing this, is to not break otherwise correct declarations just because a cv-qualifier snuck in the back door.