C++ changing multi-Dimension Array to one-D array by memory address

Question

I read in C++ book, that "Even though C++ enables us to model multidimensional arrays, the memory where the array is contained is one-dimensional. So the compiler maps the multidimensional array into the memory space, which expands only one direction."

And I thought, if I want to change n-dimension Array to 1-dimension Array(in order of Ascending) why just I call n-d array memory address(which is stored in 1-d format) and just put into new 1-d Array??

(usually I used for statement to change n-d array into 1-d array)

Is it possible?? and if it possible it would be thankful to provide simple example code.

Possible duplicate of https://stackoverflow.com/questions/48122708/is-using-2-dimensional-array-as-1-dimensional-array-correct-may-cause-undefined/48122822#48122822 — Galik, Apr 14 '18 at 07:58

score 1 · Accepted Answer · answered Apr 14 '18 at 07:00

1

A pointer to the first item in the array is also a pointer to the complete array considered as a one-dimensional array. The guarantee that the items are stored contiguously in memory comes from the sizeof requirements, that the size of an array of n items is n times the size of the item, also when the item is itself an array.

answered Apr 14 '18 at 07:00

Cheers and hth. - Alf

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And is it language conformant to simply say e.g. `static int arr3d[3][4][5]; int *p = (int *)arr3d; for(int i=0; i<3*4*5; i++) p[i] = i;` ? – Peter - Reinstate Monica Apr 14 '18 at 07:12
@PeterA.Schneider: formally not without a lot of supporting argumentation that not everybody will be convinced by, because you have a `reinterpret_cast` in there. That cast is completely unnecessary. You can just do `&arr3d[0][0][0]`. – Cheers and hth. - Alf Apr 14 '18 at 07:16
I understand what you are saying. Then for example, if there is Arr2d[2][2] then the address of Arr2d is same as address of Arr2d[0][0] so I should just input the Arr1d[4] by pointer incresement of Arr2d[0][0]. Am I right? – TaeHyeong Kim Apr 14 '18 at 07:17
@TaeHyeongKim: I think you're right, when s/should/can/ ;). – Cheers and hth. - Alf Apr 14 '18 at 07:18
@Alf well, I'm confident about the cast. It also bothers me if people dereference only in order to take the address ;-) (it's equivalent to writing `p = &(***arr3d);`) -- if anything I'd write `p = arr3d[0][0];`... – Peter - Reinstate Monica Apr 14 '18 at 07:18

score 1 · Answer 2 · answered Apr 14 '18 at 07:05

1

Yes, it is possible to convert an n-dimensional array to 1-d array.

Let me show by taking an example of 2-d array:

for (i = 0; i < n; i++)
{
    for (j = 0; j < m; j++)
    {
        b[i * n + j] = a[i][j];
    }
}

answered Apr 14 '18 at 07:05

Abhishek Keshri

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I don't think it's the way that OP expect. – con ko Apr 14 '18 at 07:07
can you clarify, this is what I understood from the question. – Abhishek Keshri Apr 14 '18 at 07:09
I done the work just like you but I want to change just using memory address. Computer store n-d array like 1-d array in memory space and I want to use it by directly calling address by reference something like that and put into directly 1-d array that I newly announced – TaeHyeong Kim Apr 14 '18 at 07:13
See my comment to Alf's answer for a code example. – Peter - Reinstate Monica Apr 14 '18 at 07:15
@AbhishekKeshri OP said that he/she often use _for_ , so he/she want to only the *address of the first element* , i think. So i think if don't use *for statement*, i think there isn't real meaning **converting**. – con ko Apr 14 '18 at 07:20
Okay, I think I got it. Editing the post. Thanks :) – Abhishek Keshri Apr 14 '18 at 07:23

C++ changing multi-Dimension Array to one-D array by memory address

2 Answers2