How to store a 2D array in a contiguous array in C?

Question

For my assignment I have to do mathematical operations on array I'm reading in from a text file. And it has to be dynamically allocated because we don't know the size of the matrix although we do know that it will be square.So I've managed to read in the array and I've put in the code so that it knows the rows and columns at run time. But my professor wants us to use contiguous arrays and allocate the matrix like this

float* A = (float*) malloc( n * m * sizeof(float) );

But I don't know how to read in the values from the text file into this array. What would I do use a for loop or something? And how would I access the elements in the array later on?

Curious, did the same professor suggest coding the `(float*)` cast? — chux - Reinstate Monica, Apr 11 '18 at 20:24
Yes the line I posted above was exactly the way he wants us to allocate it. — alex, Apr 11 '18 at 20:25
To allocate memory for a 2D array, with C99, code could use `float (*A)[n][m] = malloc(sizeof *A);` — chux - Reinstate Monica, Apr 11 '18 at 20:26
But allocation isn't the issue. He wants us to do it this way. Would doing it your way make reading it in any easier? — alex, Apr 11 '18 at 20:27
@dasblinkenlight The 1980s are calling the prof and want his compiler back. — chux - Reinstate Monica, Apr 11 '18 at 20:27
Do it the way the prof request, just know it is not up-to-date. — chux - Reinstate Monica, Apr 11 '18 at 20:28
@BobCarter Doing it chux's way would make reading it look not just easier, it would make it look completely natural, as you would write `A[r][c] = ...` and that's that. — Sergey Kalinichenko, Apr 11 '18 at 20:28
@dasblinkenlight Not sure if `A[r][c]` works. Is not `(*A)[r][c]` needed with `float (*A)[n][m] = ...`? — chux - Reinstate Monica, Apr 11 '18 at 20:46
@chux Right, of course! It's still a lot better than writing helpers that do math explicitly, though. — Sergey Kalinichenko, Apr 11 '18 at 20:48
@chux `(A*)[m] = malloc(sizeof *A * n);` will work too with `A[i][j]` it's more readable. — Stargateur, Apr 11 '18 at 20:56
@Stargateur Agreed. That is more for a dynamic 2D array than a fixed 2D array - which is usually wanted anyways. — chux - Reinstate Monica, Apr 11 '18 at 21:12

Sergey Kalinichenko · Answer 1 · 2018-04-11T20:45:57.100

5

If you must "fold" a 2D matrix into a "flat" array, use the allocation line that your professor suggested (after removing the unnecessary cast) and make a pair of functions for accessing the matrix:

float get(const float *m, size_t width, size_t r, size_t c) {
    return m[r*width+c];
}
void set(float *m, size_t width, size_t r, size_t c, float val) {
    m[r*width+c] = val;
}

Note: The above illustrates packing the matrix in row-major order; column-major order is another possibility.

edited Apr 11 '18 at 20:45

answered Apr 11 '18 at 20:34

Sergey Kalinichenko

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1

@chux Thanks! I like your `float (*A)[n][m] = malloc(sizeof *A);` "magic" a lot better, though. – Sergey Kalinichenko Apr 11 '18 at 20:46

chux - Reinstate Monica · Answer 2 · 2018-04-11T20:48:35.770

Given an allocation of the old-school form

float* A = (float*) malloc( n * m * sizeof(float) );

To read data into it

if (A) {
  for (size_t ni = 0; ni < n; ni++) {
    for (size_t mi = 0; mi < m; mi++) {
      if (fscanf(inf, "%f", A + ni*m + mi) != 1) Handle_Bad_Read();
      // or 
      if (fscanf(inf, "%f", &A[ni*m + mi]) != 1) Handle_Bad_Read();
    } 
  }

To print it

  for (size_t ni = 0; ni < n; ni++) {
    for (size_t mi = 0; mi < m; mi++) {
      printf(inf, " %g", A[ni*m + mi]);
    }
    putc('\n'); 
  }
}

score 1 · Answer 3 · answered Apr 11 '18 at 21:11

@dasblinkenlight's answer is a good one. Just to elaborate a little on how you might have figured out the math in that answer on your own: when I need to determine how to represent the relationship between two quantities mathematically, I often find it helpful to draw a picture. Excel is your friend in this case. I'll start out with a specific example. For instance, suppose I have an array with five rows and four columns. As @dasblinkenlight pointed out, there are two clear ways you might number the cells in such an array: row-major order or column-major order. I'd number them manually in my specific example:

Then I'll look at what I've drawn and try to figure out the patterns. In the first image you can see that cell numbers increase by one as you move from one column to the next, and by four (the number of columns) as you move from one row to the next. In the second image it's the opposite. So then I'll try to write a generalized example, with M rows and N columns:

Compare what I've found in my highlighted cells to @dasblinkenlight's code.

You won't always need to draw out everything by hand in order to infer the answer—the more experience you accumulate as a developer, the easier it will be to see these patterns in your head—but it's a useful approach to problems like these.

score 0 · Answer 4 · answered Apr 11 '18 at 20:40

The first part of your question is a bit unclear. As for the second part , if you are allocating a 2D matrix using contiguous allocation , you can access the values stored in it like :

 matrix[ i * <no_of_columns> + j]

where "no_of_columns" is the width of the 2D matrix , which corresponds to

 matrix[i][j]

if put simply.

How to store a 2D array in a contiguous array in C?

4 Answers4