Segmentation Fault while allocating memory at run time using malloc [duplicate] - c

Question

This question already has answers here:
How do I work with dynamic multi-dimensional arrays in C?
(9 answers)
Closed 8 years ago.
This is my code. My purpose is to allocate memory to a 2D array at run time upto whatever size is given in input.
Why is segmentation fault occuring? Is it because array elements have to be stored consecutively and malloc(dynamic allocation) is not letting this happen?
OR I am doing some error in writing this code. Please guide me through.
Thanks in advance.
int main(){
// STEP 1
int size,**arr,i=0,j=0;
printf("Enter the size of matrix : ");
scanf("%d",&size);
// STEP 2
arr = (int**)malloc(size*size*sizeof(int));
printf("\n Enter the %d elements : \n",size*size);
for(i=0;i<size;i++){
for(j=0;j<size;j++){
// STEP 3
scanf("%d",&arr[i][j]);
}
}
/*
for(i=0;i<size;i++){
for(j=0;j<size;j++){
printf("%d\n",matrix[i][j]);
}
}
*/
return 0;
}

Answer 1

This is a classic mistake.
A pointer to pointers is actually not the same as a two-dimensional array.
Granted, you can access elements of both via the var[x][y] syntax, but the memory layout of
int foo[x][y]
is different from
int **bar
If you really want to have this dynamic, you will have to allocate space for your list of pointers, then allocate space for your elements to each pointer in turn.
bar = malloc( x * sizeof(int*) );
for ( int i = 0 ; i < x ; i++ )
bar[i] = malloc( y * sizeof(int) );
If at all possible, you should try and avoid this in favor of an actual two-dimensional array, which as of C99 you can declare on the stack even if its size is determined at runtime:
int main()
{
int n;
scanf("%d", &n);
int array[n][n];
// ...
return 0;
}

Answer 2

You should allocate like this:
arr = malloc(size * sizeof(int*));
for (int i = 0; i <size; i++)
arr[i] = malloc(size * sizeof(int));
And do not forget to free the memeory using free.
Side Note: Do not cast the return value of malloc.

Answer 3

One idea that would work and would also get you rid of the memory fragmentation induced by this double pointers level allocation is make your matrix linear:
arr = (int*) malloc (size*size*sizeof(int));
And then simply access your elements with arr[i*size + j] instead of arr[i][j]:

Answer 4

use it like this : a perfect example of dynamic memory Allocation
void mxmult()
{
int n,m,a,b,c,d, sum=0;
int x,y,z;
printf("Enter first order [n*n]\n");
scanf("%d", &n);
printf("Enter second order [m*m]\n");
scanf("%d", &m);
if (n!=m)
{
printf("Invalid orders");
}
else
{
//mem allocate for matrix 1
int **mat1 = (int**)malloc(n*sizeof(int));
for(x=0;x<n;x++)
{
mat1[x]=(int*)malloc(n*sizeof(int));
}
// input matrix 1
printf("Enter the first matrix entries\n");
for (a = 0; a <n; a++)
{
for (b = 0; b < n; b++)
{
scanf("%d", &mat1[a][b]);
}
}
// memory allocate matrix 2
int **mat2 = (int**)malloc(m*sizeof(int));
for(y=0;y<n;y++)
{
mat2[y]=(int*)malloc(m*sizeof(int));
}
//inpur matrix 2
printf("Enter the second matrix entries\n");
for (c = 0; c <n; c++)
{
for (d= 0; d < n; d++)
{
scanf("%d", &mat2[c][d]);
}
}
//Memory allocate matrix Mult
int **mult=(int**)malloc(m*sizeof(int));
for(z=0;z<m;z++)
mult[z]=(int*)malloc(m*sizeof(int));
for (a = 0; a < n; a++)
{
for (d = 0; d < m; d++)
{
for (c = 0; c < n; c++)
{
sum=sum + (mat1[a][c] *mat2[c][d]);
}
mult[a][d] = sum;
sum= 0;
}
}
printf("Product\n");
for ( a = 0 ; a < n ; a++ )
{
for ( d = 0 ; d < m ; d++)
printf("%d\t", mult[a][d]);
printf("\n");
}
}
}