Const qualifier ignored [duplicate]

Question

This question already has answers here:

typedef pointer const weirdness (6 answers)

Closed 5 years ago.

I have a struct type_s. Then I typedef a pointer to a struct type_s as type.

If I have a const struct type_s* then the compiler will correctly complain if an assignment is made to the struct member, as seen in function Test1(). But if I make a const type, which is a typedef for the same struct pointer the compiler will not complain and compile, function Test2().

typedef struct type_s* type ;
struct type_s
{
    int a ;
} ;

void Test1( const struct type_s* t )
{
    t->a = 123 ;  //complains, ok
}

void Test2( const type t )
{
    t->a = 123 ;   //doesn't, it should in my oppinion
}

To me they are logically both the same thing. What am I missing?

Is the only solution to create another typedef for a constant pointer to that struct like this:

typedef const struct type_s* ctype ;

which will work correctly as in Test1().

Answer 1

What you're missing is that a const pointer is not the same thing as a pointer to a const object. You have one of each. A const pointer is a pointer whose stored address is not modifiable; a pointer to const is one which cannot be used to modify its referent.

Answer 2

They are not the same thing. const has different semantics when applied to the definition or declaration of a pointer variable and a non-pointer variable.

const struct type_s *t 

The above statement defines t to be a pointer to an object of type const struct type_s. Here, const qualifies the type struct type_s of the object t points to, not the pointer t.

typedef struct type_s *type ;

The above statement defines an alias named type for the type struct type_s *.

const type t;
// equivalent to
type const t;
// equivalent to
struct type_s *const t;

The above statement defines t to be a constant object of type type. Here const qualifies the object. You cannot replace the typedef with its original type and reinterpret it as if it were a macro. The pointer information is embedded in the typedef and it will be treated as a non-pointer type.