Question

Is it true that C++ array is a const pointer in C++ ?

http://www.cplusplus.com/forum/articles/10/#msg1381

So why is x and &x the same if they are pointers?

#include<iostream>
using namespace std;

int main() {
    int x[5];
    int *y = new int[5];

    cout << "&x : " << &x << endl;
    cout << "x : " <<  x << endl;
    cout << endl;
    cout << "&y : " << &y << endl;
    cout << "y : " <<  y << endl;
    return 0; 
}

Output:

&x:  0xbfec4ccc
x :  0xbfec4ccc

&y:  0xbfec4ce0
y :  0x8961008

Link to ideone of above code:
http://ideone.com/3JRZd

Answer 1

No, it is not true that an array is a const pointer in C++. Indeed if one file contains

extern int* const a;

and the other contains

int a[10];

in the same scope, combining both files will cause strange errors.

However, arrays decay into pointers in most situations. More exactly, if you use an array in an rvalue context, it is implicitly converted into a pointer to its first element. This is what happens to x in the example in the second link: Since the first element of the array of course is at the same address as the array, both x and &x are printed as the same address. However note that they are not the same pointer because they are of different types: x in this context decays into int*, i.e. pointer to int, while &x is of type int(*)[5], i.e. pointer to array of 5 ints.

Another case where you can see that both are different is a function taking a reference to an array, like

void f(int (&arr)[5]);

You can pass x to this function (because it is of type int[5]) but neither &x nor the pointer y you get when writing int* y = x;