Simulate 2D array to 1D

Question

I have a jagged 2D array, the rows aren't all the same length:

int sizes[100];
//init sizes

unsigned char *p = [100];

for(unsigned int i = 0; i < 10; i++)
{
    p[i] = (unsigned char*)malloc(sizeof(char)*sizes[i]);
    for(unsigned int j = 0; j < sizes[i]; j++)
        p[i] = j;
}

I use the array like this:

p[x][y]

How can I simulate this array to 1D?

If you have a "jagged array" (different lengths), then it is really not advisable to try to make it a one D array. How did you create it? — Floris, Dec 24 '13 at 03:12
I have no other options, the lengths must be different, what's faster for this job, other than a 2d array? — Blazer, Dec 24 '13 at 03:20
Why use `malloc` and tag the question C++? Why not `new` or better still vectors — Ed Heal, Dec 24 '13 at 03:25
I do not believe the duplicates referenced are relevant since this refers to a jagged array (not mxn). — Floris, Dec 24 '13 at 16:42

Lawrence Jones · Answer 1 · 2013-12-24T03:20:33.887

2

If you're looking for a way to map a two dimensional array onto a one dimensional space, then try...

int sizes[width*height];
void setPoint(int x, int y, int val) {
  sizes[x*width + y] = val;
}

Notably the x*width + y indexing will give you the appropriate element in the one dimensional array.

edited Dec 24 '13 at 03:20

answered Dec 24 '13 at 03:06

Lawrence Jones

955
6
14

if the size of the 2nd dimension on each array is different, is it possible to use this? – Blazer Dec 24 '13 at 03:08
@Blazer - yes, if you store the dimension somewhere, then you just sum the first x elements instead of multiplying. – Paweł Stawarz Dec 24 '13 at 03:21

score 2 · Accepted Answer · answered Dec 24 '13 at 03:58

I assume that if you want to access your "2D array" as a one D matrix, you expect that as you increment your index by 1, you access the next element in the array (and will automatically go to the next line when you run off the edge). The right way to do this is by changing the way you allocate the array. I'll try to show how this is done - this is just C, not C++. It's probably more appropriate since you were using malloc anyway. I am also thinking that you have a pretty serious error in your code, in that you are creating a char * pointer p, yet expect to use it in

p[x][y];

for which you would need a char **, obviously. Let's try to make code that would do what you want:

int sizes[100];
//init sizes

unsigned char *p[100];  // without the == sign we create an array of 100 pointers
unsigned char *bigP;   // the master pointer

int totalLength = 0;

int ii;
for(ii=0; ii<100; ii++) totalLength += sizes[ii];

bigP = malloc(sizeof(char) * totalLength);

int offset = 0;

// make pointers p point to places along this big memory block:
for(ii = 0; ii < 100; ii++) {
  p[ii] = bigP + offset;
  offset += sizes[ii];
}

Now you can either address your array with

p[x][y];

or with

bigP[z];

where z can go from 0 to the maximum number of elements. Of course, in the process you don't know (when you are using the "1D" paradigm) in what row/column of the jagged array you are - you can't know that if you are truly in one dimension.

The key here is that the memory is allocated as a single contiguous block, and the pointers p are pointing to places in that block. This means you MUST NOT FREE p. You must only ever free bigP.

I hope this makes sense.

score 1 · Answer 3 · answered Dec 24 '13 at 03:29

1

Unless it is homework just download boost and use Boost.Matrix.

answered Dec 24 '13 at 03:29

Öö Tiib

10,809
25
44

Simulate 2D array to 1D

3 Answers3