How can I allocate a matrix with structs?

Question

I must write a function that should help me to allocate a matrix using a struct. I Began to study the structs today. So I wrote this code with that struct and relative main to prove the function:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

typedef struct {

  int rows; 
  int cols;
  float **row_ptrs;
} Mat; 

Mat* Mat_alloc(int rows, int cols);

int main(int argc, char **argv) 
{
      Mat *m1 = Mat_alloc(int rows, int cols);

     return 0;
}
Mat* Mat_alloc(int rows, int cols)
{
    Mat matrice;
    matrice.rows = rows;
    matrice.cols = cols;
    float** matrice= (float**)malloc((matrice.rows)*sizeof(float*));
    for(int i = 0; i < matrice.cols; i++)
    {
        matrice[i] = (float*)malloc((matrice.cols)*sizeof(float));
    }
    matrice.row_ptrs = matrice;
    return matrice;
}

I know that I make some mistakes.Someone can help me to undestand how I can do it ?

First find a [good beginners book](http://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list) and learn how to call functions. Then continue reading the books and learn about *scopes* and what happens with local variables once the function they were defined in returns. — Some programmer dude, Apr 29 '16 at 14:16
Note: Code will eventually need with a companion `Mat_free(Mat *);` — chux - Reinstate Monica, Apr 29 '16 at 16:23
There is no matrix (aka 2D array) in your code and nothing which can be used as one. A pointer is not an array and vice versa. — too honest for this site, May 12 '16 at 15:16

score 2 · Answer 1 · 2016-04-29T14:22:21.547

2

int rows and int cols are uninitialized going into Mat_alloc. You need to give those numerical values!

int main(int argc, char **argv) 
{
      int rows = 10;
      int cols = 10;
      Mat *m1 = Mat_alloc(rows, cols);

      //do something
      //call your Mat_free(m1) function

     return 0;
}

Make sure you return the pointer to a Mat structure in this function too:

Mat* Mat_alloc(int rows, int cols)
{
    Mat *m1 = malloc(sizeof(Mat));
    m1->rows = rows;
    m1->cols = cols;
    float** matrice= (float**)malloc((m1->rows)*sizeof(float*));
    for(int i = 0; i < m1->rows; i++)
    {
        matrice[i] = (float*)malloc((m1->cols)*sizeof(float));
    }
    m1->row_ptrs = matrice;
    return m1;
}

Also, make sure you create a Mat_free function to free up the memory allocated in Mat_alloc.

edited Apr 29 '16 at 14:22

answered Apr 29 '16 at 14:16

Thank you for your comment! – mikecolux95 Apr 29 '16 at 14:45
Curious: why the unnecessary cast `(float*)` in `matrice[i] = (float*)malloc((m1->cols)*sizeof(float));`? – chux - Reinstate Monica Apr 29 '16 at 14:46
@chux naughty question ;) – LPs Apr 29 '16 at 14:49
I use this because the function malloc is a void* function. – mikecolux95 Apr 29 '16 at 14:52
@LPs Yes and no. I _am_ interested in why folks use certain coding paradigms that have debatable value. – chux - Reinstate Monica Apr 29 '16 at 14:53
@MicheleColuccelli but you are coding using [tag:c] where it is allowed to assign a `void *` to another type, rather then [tag:c++] where is not allowed. – LPs Apr 29 '16 at 14:55
@MicheleColuccelli Take a look at [this fmaous post](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc) – LPs Apr 29 '16 at 14:56
@MicheleColuccelli Another thing that is not necessary is to use a temp variable (`matrice`) to create the matrix. You could use `m1->row_ptrs` directly. – LPs Apr 29 '16 at 15:01
@LPs Thank you for this! – mikecolux95 Apr 29 '16 at 15:03
@MicheleColuccelli Prego ;) – LPs Apr 29 '16 at 15:06
@chux I try to leave the code as similar to the OP's code as possible, but in working order. I don't want to reinterpret his motives for casting / using tmp variables. – Apr 29 '16 at 15:42
Agree, Not fixing working legacy code is a reason to leave the cast. – chux - Reinstate Monica Apr 29 '16 at 16:25

score 1 · Accepted Answer · answered Apr 29 '16 at 15:06

1

chrisd1100 gave a good answer, but only to be a little bit pedantic this is mine:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

typedef struct {

  int rows;
  int cols;
  float **row_ptrs;
} Mat;

Mat* Mat_alloc(int rows, int cols);

int main(void)
{
    int i;

    int rows = 10;
    int cols = 10;

    Mat *m1 = Mat_alloc(rows, cols);

    for (i=0; i<cols; i++)
    {
        free(m1->row_ptrs[i]);
    }
    free(m1->row_ptrs);
    free(m1);

    return 0;
}

Mat* Mat_alloc(int rows, int cols)
{
    Mat *m1 = malloc(sizeof(Mat));
    m1->rows = rows;
    m1->cols = cols;
    m1->row_ptrs = malloc((m1->rows)*sizeof(float*));
    for(int i = 0; i < m1->rows; i++)
    {
        m1->row_ptrs[i] = malloc((m1->cols)*sizeof(float));
    }

    return m1;
}

answered Apr 29 '16 at 15:06

LPs

16,045
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let me see if I really known how to deallocate it. I had to do, first two deallocations with that I can deallocate the matrix defined in the structs Mat (float **row_ptrs), but then I had also to deallocate m1 that is of type Mat. Isn' it? – mikecolux95 Apr 29 '16 at 15:23
Yes, it is. You free them in the reverse order you allocate them. – LPs Apr 29 '16 at 15:29
Thank you for your support :) – mikecolux95 Apr 29 '16 at 15:41
Ri-prego. Di nulla – LPs Apr 29 '16 at 15:42

How can I allocate a matrix with structs?

2 Answers2