C ++内存和指针地址

Question

I am reading a great book by Stephen Prata. Unfortunately I don't understand something. It is about array and pointers. So there is array tab. He wrote, that tab is the address for first element of array -> that is ok, but &tab is address for whole array <- I don't understand this. What does it means? &tab shows all address of array (of course not), middle address of all elements of array, the last one?

Answer 1

In C and C++ there are two aspects to a pointer:

• The memory address it points to
• The type of what it is pointing to

tab and &tab indicate the same address, but they have different type. They denote different objects: &tab is a large array, and tab (aka. &tab[0] ) is a sub-object of that array.

It's no different to this situation:

struct S
{
    int x;
    int y;
};

S s;

S *p1 = &s;
int *p2 = &s.x;

In this case p1 and p2 both point to the same address, but they have different types, and point to different objects which occupy the same space.

Answer 2

tab is the address for first element of array

this is correct, for char arr[1]; you can write char* p = arr; , here we say that arr decays to pointer to its first element.

&tab is address for whole array

it means its type is a pointer to array, ie. char(*)[1] . Their pointer values (arr and &arr) are equal but the types differ. So below will compile:

char (*pp)[1] = &arr; // pointer to array
char (&rp)[1] = arr; // reference to array

but this will not compile:

char* pp = &arr; // error, types differ

You can use below trick to find what type given expression/variable really is. Compiler will show you an error from which you can read actual types:

template <typename T>
struct TD;

int main()
{
char arr[1];

//char* pp = &arr; // error, types differ

TD<decltype(arr)> dd;    // is char[1], no decay to char* here
TD<decltype(&arr[0])> dd;    // is char*
//TD<arr> dd;     // char(*)[1]
}