Dynamically 2D array in C using the single pointer: - c

Question

I'm trying to dynamically allocate memory to a 2d array using a single pointer. For that, I have 3 functions that allocate the respective memory newarray() and to store individual elements in it store(), to fetch elements from it fetch(). I don't know why I get execution errors while I test it, also I should allocate the the exact amount of memory for it, that might be the problem but I'm not sure how to do that. This probrams deals with a triangular matrix which should have the number of columns lower than the number of rows when It comes to adding elements, like I, have a 5x5 array where (4,2) and (4,4) its OK but (4,5) its NOT.
Here is the code
typedef int* triangular;
triangular newarray(int N){
triangular mat = NULL; //pointer to integer
//Allocate memory for row
mat = (int *)malloc(N * N * sizeof(int));
//Check memory validity
if(mat == NULL)
{
return 1;
}
return mat;
}
int store(triangular as, int N, int row, int col, int val){
if(row >= col){
as[row * N + col] = val;
return 1;
}else if(row < col){
return -1;
}else if((row > N) ||(col > N) || (row + col > N + N))
return -1;
}
int fetch(triangular as, int N, int row, int col){
int value;
value = as[row * N + col];
if((row > N) ||(col > N) || (row + col > N + N) )
return -1;
else if(row < col)
return -1;
return value;
}
nt main()
{
int iRow = 0; //Variable for looping Row
int iCol = 0; //Variable for looping column
int N;
triangular mat = newarray(5);
printf("\nEnter the number of rows and columns = ");
scanf("%d",&N); //Get input for number of Row
store(mat,N,3,2,10);
store(mat,N,3,3,10);
store(mat,N,4,2,111);
store(mat,N,3,5,11);
printf("the element at [3,5] is : %i", fetch(mat,N,3,5));
//Print the content of 2D array
for (iRow =0 ; iRow < N ; iRow++)
{
for (iCol =0 ; iCol < N ; iCol++)
{
printf("\nmat[%d][%d] = %d\n",iRow, iCol,mat[iRow * N + iCol]);
}
}
//free the allocated memory
free(mat);
return 0;
}

Answer 1

int store(triangular as, int N, int row, int col, int val){
if(row >= col){
as[row * N + col] = val;
return 1;
}else if(row < col){
return -1;
}else if((row > N) ||(col > N) || (row + col > N + N))
return -1;
}
in store function, first if condition is so weird. Why you dont set the value to the array when the parameters passed to function is 2(row), 3(column).
I changed your store in the following way. index and array size are different things because of that index is equal to N - 1. In your code, there are a lot of if checks I guess checking only row and col is enough to understand that they are inside boundaries.
int store(triangular as, int N, int row, int col, int val){
int index = N - 1;
if((row > N) ||(col > N))
return -1;
as[row * index + col] = val;
return 1;
}
I changed your fetch function like below because the reason I mentioned about your store function.
int fetch(triangular as, int N, int row, int col){
int value;
int index = N - 1;
if((row > index) ||(col > index))
return -1;
value = as[row * index + col];
return value;
}

Answer 2

You are making this needlessly complicated. All those functions and manual run-time calculations aren't really necessary.
Also, you have the following problems:
Don't hide pointers behind typedef, it just makes the code unreadable for no gain.
Initialize the data returned from malloc or instead use calloc which sets everything to zero, unlike malloc.
Arrays in C are zero-indexed so you can't access item [3][5] in an array of size 5x5. This is a common beginner problem since int array[5][5]; declares such an array but array[5][5] for index accessing goes out of bounds. The syntax for declaration and access isn't the same, access needs to start at 0.
You didn't include any headers, I'm assuming you left that part out.
Here's a simplified version with bug fixes that you can use:
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int N=5;
int (*mat)[N] = calloc( 1, sizeof(int[N][N]) ); // allocate 2D array dynamically
if(mat == NULL)
return 0;
mat[3][2] = 10;
mat[3][3] = 10;
mat[4][2] = 111;
mat[3][4] = 11;
for(int i=0; i<N; i++)
{
for(int j=0; j<N; j++)
{
printf("[%d][%d] = %d\n", i, j, mat[i][j]);
}
}
free(mat);
return 0;
}
Further study: Correctly allocating multi-dimensional arrays