int a[2][2] = {{1, 2}, {3, 4}};
    cout << a << endl;
    cout << *a << endl;
    cout << &a[0][0] << endl;

The output of this code is:

     0x7fff3da96f40

     0x7fff3da96f40

     0x7fff3da96f40

However, if a is 0x7fff3da96f40 then *a should be the value in the address 0x7fff3da96f40 which is 1.

But, we get the same address again.

Why is this happening?

CodePudding user response：

*a should be the value in the address 0x7fff3da96f40

It is. The key is what type the value has, and that type is int[2].

cout can't print arrays directly, but it can print pointers, so your array was implicitly converted to a pointer to its first element, and printed as such.

Most uses of arrays do this conversion, including applying * and [] to an array.

CodePudding user response：

If you try

int a[2][2] = {{1, 2}, {3, 4}};
cout << a << endl;
cout << *a << endl;
cout << &a[0][0] << endl;
cout << typeid(a).name() << endl;
cout << typeid(*a).name() << endl;
cout << typeid(&a[0][0]).name() << endl;

the possible output is

0x7ffe4169bb50
0x7ffe4169bb50
0x7ffe4169bb50
A2_A2_i
A2_i
Pi

So now you can see what types are the variables you are using. They are somewhat different entities, starting on same memory address.