Multi dimensional array and pointer in c++

Question

    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?

Try *a = 0 and you see in the error message what *a is. This is not int. — 3CEZVQ
– 3CEZVQ, Commented Oct 14, 2021 at 7:14
a is int (&)[2][2], *a is int(&)[2], a[0][0] is an int&. The first two decay to pointer. — Jarod42
– Jarod42, Commented Oct 14, 2021 at 7:59
It happens because a isn't 0x7fff3da96f40 – ais not a pointer but an array. The value you see is its decay into a pointer to its first element, &a[0] (and that first element is also an array that decays into a pointer to its first element when you print it). — molbdnilo
– molbdnilo, Commented Oct 14, 2021 at 8:09
@Jarod42 Excuse my nitpicking, but they aren't references. Expressions can't have reference types. — HolyBlackCat
– HolyBlackCat, Commented Oct 14, 2021 at 8:11
Very often, when you think "if A is true, then B should happen, so why doesn't B happen?", it is because A isn't true. — molbdnilo
– molbdnilo, Commented Oct 14, 2021 at 8:16

HolyBlackCat · Accepted Answer · 2021-10-14 07:15:23Z

3

*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.

answered Oct 14, 2021 at 7:15

HolyBlackCat

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Comments

ygroeg · Accepted Answer · 2021-10-14 07:30:02Z

0

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.

answered Oct 14, 2021 at 7:30

ygroeg

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