Double Dynamic Array

Question

This is my first time working with dynamic arrays and I am honestly just so confused. My original array(below) works fine.

#include <stdio.h>

int main()
{
    char icdC[4][10];

    for(int i =0;i<4;i++){
      for(int j=0;j<10;j++){
        printf("What are your ICD codes [%d][%d]\n",i,j);
        scanf("%s", &icdC[i][j]);
      }
    }

    return 0;
}

However I tried converting this array into a double dynamic array (below) it doesn't seem to work correctly as it will either tell me "signal: segmentation fault (core dumped) or it just won't run.

#include <stdio.h>
#include <strings.h>
#include <stdlib.h>
#include <ctype.h>
#include <stdbool.h>

int main()
{

    char** icdC;
    icdC = (char**)malloc(4*10*sizeof(char));


    for(int i=0;i<4;i++){
      for(int j=0;j<10;j++){
        printf("What are your ICD codes [%d][%d]\n",i,j);
        scanf("%s", &icdC[i][j]);
      }
    }

    return 0;
}

Answer 1

The issue seems to be located within icdC = (char**)malloc(4*10*sizeof(char));, as it is an incorrect reference to the second array you're trying to create.

In C there are many options to represent a 2d array, among which the possibility to create an array of pointers, each of which points to a specific row. Then you can allocate the memory for each array using the malloc, which allows you to generate an "array of pointers".

The implementation would most likely be:

char ** icdC;
int r = 4, c = 10;
icdC = (char*) malloc(sizeof(char*) * r);
for (int i = 0; i < r; i++) {
    icdC[i] = malloc(sizeof(char) * c);
}

Note: You can also do this implementation in two loops as you did above.

From here you can do a nested loop to scan the value like you've already done in your snippet. Remember that now you can access a value by indicating the index for the corresponding row and column, like this: icdC[row_num][row_column] = .....

It might look strange at the beginning, but after you get a grasp of the concept, it's all good and fun!