C - How to adapt this code for a very large/dynamic array to a 2 dimensional array?

Question

 int main()
  {

    double *array;
    long int n;
    n=10000000;//10^7
    array = (double *) malloc(n*sizeof(double));
    return 0;
  }

basically, I want to use this code for a really big aray into a 2 dimensional array, which will have dimensions [very large][4].

Answer 1

If you want a 2D array, then allocate a 2D array. It's that simple.

double (*pArr)[4] = malloc(10000000 * sizeof pArr[0]);

Notes:

• do not cast the return value of malloc().
• use sizeof pArr[0] instead of sizeof(TheDataType) for defensive programming reasons.

Answer 2

This seems working on Wandbox.

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

int main(void) {
    double (* array)[4];
    long int n;
    int i, j;
    n=10000000;//10^7
    array =  (double (*)[4]) malloc(n*sizeof(double[4]));

    printf("%u\n",(unsigned int)sizeof(array[0]));
    printf("%u\n",(unsigned int)sizeof(double[4]));
    for (i = 0; i <n; i++) {
        for (j = 0; j < 4; j++) array[i][j] = (double)i * j;
    }
    for (i = 0; i < 10; i++) {
        for (j = 0; j < 4; j++) printf("%f ", array[i][j]);
        putchar('\n');
    }
    for (i = n - 10; i < n; i++) {
        for (j = 0; j < 4; j++) printf("%f ", array[i][j]);
        putchar('\n');
    }
    free(array);
    return 0;
}

Answer 3

int n = 100000;
double** array = malloc(sizeof(double*)*n);
for (int i = 0; i < n; ++i)
{
    array[i] = malloc(4*sizeof(double));
}

Also note that we don't cast the malloc's result(Do I cast the result of malloc?).