C&C++中带有const限定符的指向数组的指针

Question

Consider following program:

int main()
{
    int array[9];
    const int (*p2)[9] = &array;
}

It compiles fine in C++ (See live demo here ) but fails in compilation in C. By default GCC gives following warnings. (See live demo here ).

prog.c: In function 'main':
prog.c:4:26: warning: initialization from incompatible pointer type [enabled by default]
     const int (*p2)[9] = &array;

But If I use -pedantic-errors option:

gcc -Os -s -Wall -std=c11 -pedantic-errors -o constptr constptr.c

it gives me following compiler error

constptr.c:4:26: error: pointers to arrays with different qualifiers are incompatible in ISO C [-Wpedantic]

Why it fails in compilation in C but not in C++? What C & C++ standard says about this?

If I use const qualifier in array declaration statement it compiles fine in C also. So, what is happening here in above program?

Answer 1

GCC-gnu

In GNU C, pointers to arrays with qualifiers work similar to pointers to other qualified types. For example, a value of type int (*)[5] can be used to initialize a variable of type const int (*)[5] . These types are incompatible in ISO C because the const qualifier is formally attached to the element type of the array and not the array itself .

C standard says that (section: §6.7.3/9):

If the specification of an array type includes any type qualifiers, the element type is so- qualified, not the array type .[...]

Now look at the C++ standard (section § 3.9.3/5):

[...] Cv-qualifiers applied to an array type attach to the underlying element type, so the notation “ cv T ,” where T is an array type, refers to an array whose elements are so-qualified. An array type whose elements are cv-qualified is also considered to have the same cv-qualifications as its elements . [ Example :

 typedef char CA[5]; typedef const char CC; CC arr1[5] = { 0 }; const CA arr2 = { 0 };

The type of both arr1 and arr2 is “array of 5 const char,” and the array type is considered to be const- qualified . —endexample]

Therefore, the initialization

const int (*p2)[9] = &array;  

is assignment of type pointer to array[9] of int to pointer to array[9] of const int . This is not similar to assigning int * to a const int * where const is applied directly to the object type the pointer points to . This is not the case with const int(*)[9] where, in C, const is applied to the elements of the array object instead of the object the pointer points to. This makes the above initialization incompatible.

This rule is changed in C++. As const is applied to array object itself, the assignment is between same types pointer to const array[9] of int instead of type pointer to array[9] of int and pointer to array[9] of const int .