C & C++: What is the difference between pointer-to and address-of array?

Question

C++11 code:

int a[3];
auto b = a;       // b is of type int*
auto c = &a;      // c is of type int(*)[1]

C code:

int a[         


        

Answer 1

The sizeof operator should behave differently, for one, especially if you change the declaration of a to a different number of integers, such as int a[7]:

int main()
{
    int a[7];

    auto b = a;
    auto c = &a;

    std::cout << sizeof(*b) << std::endl;  // outputs sizeof(int)
    std::cout << sizeof(*c) << std::endl;  // outputs sizeof(int[7])

    return 0;
}

For me, this prints:

4
28

That's because the two pointers are very different types. One is a pointer to integer, and the other is a pointer to an array of 7 integers.

The second one really does have pointer-to-array type. If you dereference it, sure, it'll decay to a pointer in most cases, but it's not actually a pointer to pointer to int. The first one is pointer-to-int because the decay happened at the assignment.

Other places it would show up is if you really did have two variables of pointer-to-array type, and tried to assign one to the other:

int main()
{
    int a[7];
    int b[9];

    auto aa = &a;
    auto bb = &b;

    aa = bb;

    return 0;
}

This earns me the error message:

xx.cpp: In function ‘int main()’:
xx.cpp:14:8: error: cannot convert ‘int (*)[9]’ to ‘int (*)[7]’ in assignment
     aa = bb;

This example, however, works, because dereferencing bb allows it to decay to pointer-to-int:

int main()
{
    int a;
    int b[9];

    auto aa = &a;
    auto bb = &b;

    aa = *bb;

    return 0;
}

Note that the decay doesn't happen on the left side of an assignment. This doesn't work:

int main()
{
    int a[7];
    int b[9];

    auto aa = &a;
    auto bb = &b;

    *aa = *bb;

    return 0;
}

It earns you this:

xx2.cpp: In function ‘int main()’:
xx2.cpp:14:9: error: incompatible types in assignment of ‘int [9]’ to ‘int [7]’
     *aa = *bb;