Why *(arr+1) brings the same value as arr+1 in multi-dimensional arrays?

Question

I have a doubt about multi-dimensional arrays and pointers notation, as below.

int arr[2][4] = {{1,2,3,4}, {5,6,7,8}};
printf("%d\n", (arr+1));
printf("%d\n", *(arr+1));

Why in both printf() calls, the result printed is the same? Both bring the addresses of the second array inside of the arr. The first one I understand because arr is a memory address and so, adding 1 to it moves the pointer ahead by (the size of the inner array * size of integer). But for the second one I am confused because if I put * before (arr+1), shoudn't it be the value on the address of (arr+1)?

The code presented is not valid C. Putting a data type, maybe `int`, in front of `arr[2][4]` would fix the worst problem by making that into a declaration. — John Bollinger, Jun 23 '23 at 16:27
With additional useful related info in [How come an array's address is equal to its value in C?](https://stackoverflow.com/q/2528318/364696) and [What is array to pointer decay?](https://stackoverflow.com/q/1461432/364696). — ShadowRanger, Jun 23 '23 at 16:48
@Filipe Andrade, "result printed in the screen is the same" --> technically the 2 pointers do not need to be the _same_ bit pattern, just _equivalent_ pointers - once converted to `void *`. — chux - Reinstate Monica, Jun 23 '23 at 17:07

score 1 · Answer 1 · answered Jun 23 '23 at 16:29

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Values are the same, but types are different: int(*)[4] (pointer to array of 4 ints) vs int * (pointer to int, but before the implicit conversion caused by passing it to a function it was int[4], an array of 4 ints).

Those two values are &arr[1] and &arr[1][0] respectively. The first subarray and its first element have the same address.

answered Jun 23 '23 at 16:29

HolyBlackCat

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... and this is a many-times-over dupe, but making this comment is all I'm going to do about it because I am presently refraining from all community moderation actions. – John Bollinger Jun 23 '23 at 16:34
@JohnBollinger Can you give me a good dupe link? – HolyBlackCat Jun 23 '23 at 16:46

pmg · Answer 2 · 2023-06-23T17:08:25.413

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arr + 1 is the address of a block of 4 items.
*(arr + 1) is the address of 1 item.

edited Jun 23 '23 at 17:08

answered Jun 23 '23 at 16:24

pmg

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Wrong `arr+1` is just `&arr[1]`. – Jean-Baptiste Yunès Jun 23 '23 at 16:30
Edited, thanks @Jean-BaptisteYunès – pmg Jun 23 '23 at 17:09

score 0 · Accepted Answer · answered Jun 23 '23 at 16:35

(arr) points to the first memory address of the 2 * 4 array, i.e., arr[0][0].
(arr+1) points to the 2nd 1-D array comprising arr, i.e., it points to the array arr[1].

Now, *(arr+1) is basically the address of the first element of the 2nd 1-D array comprising arr, i.e., it points to the element arr[1][0].

If you know pointers well, arr[1] is the same address as arr[1][0].

You can confirm that by checking **(arr+1), which will output 5 in your case.

Why *(arr+1) brings the same value as arr+1 in multi-dimensional arrays?

3 Answers3