What does arr++ mean in a function "int f(int arr[])"?

Question

I was looking at this code and understood pretty much all of it except one thing:
what does Arr1++ means?

What does it do to the array? since Arr1 is not just a normal variable like int..

bool theSameElements(int Arr1[], int Arr2[], int size)
{
    int temp;
    if (size == 0)
    {
        return true;
    }
    for (int i = 0; i < size; i++)
    {
        if (Arr1[0] == Arr2[i])
        {
            temp = Arr2[i];
            Arr2[i] = Arr2[0];
            Arr2[0] = temp;
            Arr1++;
            Arr2++;
            return theSameElements(Arr1, Arr2, size - 1);
        }
    }
    return false;
}

Despite being declared as `int Arr1[]`, `Arr1` is a pointer to int, so the code is incrementing the pointer. — , Jan 07 '18 at 16:38
Worthwhile reading: [What is array decaying?](https://stackoverflow.com/questions/1461432/what-is-array-decaying) — user4581301, Jan 07 '18 at 16:49
Incrementing the address. An array name is just a pointer to its first element `arr[0]` thus to move to the next element you either increment the address or use `index`. — Raindrop7, Jan 07 '18 at 17:35

Ron · Accepted Answer · 2018-01-07T21:09:33.440

2

Any array passed as function parameter is implicitly converted / decays to a pointer of type int*. Now the Arr1 is a pointer pointing to the first array element namely Arr1[0]. This is known as the Array-to-pointer decay. Applying the post-increment operator:

Arr1++;

increments the pointer value by the size of the data it points to so now it points to the second array element Arr1[1].

That being said you should prefer std::array to raw arrays and smart pointers to raw pointers.

edited Jan 07 '18 at 21:09

answered Jan 07 '18 at 16:46

Ron

14,674
4
34
47

1

There's no decaying going on here. The result of array-to-pointer decaying is an rvalue, and `++` requires an lvalue. – Sneftel Jan 07 '18 at 17:03
See https://ideone.com/MwLWhX for what happens when actual array-to-pointer decaying happens. – Sneftel Jan 07 '18 at 17:18
Thanks. If I understood it right- when I do arr++; it means that the "new" arr[0] is actually arr[1] or ? – Cocoboom Jan 07 '18 at 17:19
As said, the parameter `Arr1` **is not** an array, but a pointer. (The argument passed *might* be, and will then decay to a pointer.) Never confuse the two, even if the syntax can be deceptive. – Deduplicator Jan 07 '18 at 17:24
I understand.Thank you – Cocoboom Jan 07 '18 at 17:37

score 1 · Answer 2 · answered Jan 07 '18 at 16:44

1

So what you need to know is that Arr, in this case, is the pointer to the array. So when you do Arr++ you are basically incrementing the pointer to this array.

answered Jan 07 '18 at 16:44

Khaldoun Nd

1,436
12
23

score 1 · Answer 3 · edited Jan 07 '18 at 17:30

1

Function parameters which appear to have array type, in fact have pointer type. int foo[3] and int foo[], in the context of a function parameter list (and only in that context), are exactly the same as int* foo. Because of this, you can do things to them that you wouldn't ordinarily be able to do to arrays, such as reassign their values.

edited Jan 07 '18 at 17:30

Deduplicator

44,692
7
66
118

answered Jan 07 '18 at 17:06

Sneftel

40,271
12
71
104

What does arr++ mean in a function "int f(int arr[])"?

3 Answers3