Allocating a 2D array using calloc - c

Question

I am trying to dynamically allocate a 2D array using calloc.
Here is the code that I have tried.
int ** numberOfConstPiArray = calloc(invariannumberOfUniqueKernels * kernelColumnCount, sizeof(int *));
I have initialised the following variables as shown
int numberOfUniqueKernels = 100;
int kernelColumnCount = 10;
int dimensionalMatrixColumnCount = 10;
The following is the main code which loops through and tries to alter the 2D array.
for (int countKernel = 0; countKernel < numberOfUniqueKernels; countKernel++)
{
int countNumberOfConst = 0;
int numberOfTerms = 0;
int numberOfConstPi = 0;
for (int col = 0; col < kernelColumnCount; col++)
{
for (int row = 0; row < dimensionalMatrixColumnCount; row++)
{
if (some condition is satisfied)
{
countNumberOfConst += 1;
}
if (another condition satisfied)
{
numberOfTerms += 1;
}
}
if(countNumberOfConst == numberOfTerms)
{
numberOfConstPi += 1;
numberOfConstPiArray[countKernel][col] = 1;
}
countNumberOfConst=0;
numberOfTerms=0;
}
}
This doesn't seem to work. I understand that doesn't seem to work is vague but as this code is a part of a large compiler, there is no way for me to print out the specific output. Apologies for that.
My question is: Have I initialised the arrays in the correct way and have is the way I modified the values of the elements in the array correct.
Thank you.

Answer 1

This
int ** numberOfConstPiArray = calloc(invariannumberOfUniqueKernels * kernelColumnCount, sizeof(int *));
is not an allocation of a two-dimensional array because at least the type of numberOfConstPiArray is int ** instead of for example int ( * )[kernelColumnCount].
If your compiler supports variable length arrays then you could use the following approach as it is shown in the demonstrative program
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
size_t n = 5;
int ( *a )[n] = calloc( n * n, sizeof( int ) );
for ( size_t i = 0; i < n; i++ )
{
for ( size_t j = 0; j < n; j++ ) a[i][j] = i * n + j;
}
for ( size_t i = 0; i < n; i++ )
{
for ( size_t j = 0; j < n; j++ ) printf( "%2d ", a[i][j] );
putchar( '\n' );
}
free( a );
return 0;
}
The program output is
0 1 2 3 4
5 6 7 8 9
10 11 12 13 14
15 16 17 18 19
20 21 22 23 24
Or you can allocate an array of arrays the following way.
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
size_t n = 5;
int **a = calloc( n, sizeof( int * ) );
for ( size_t i = 0; i < n; i++ )
{
a[i] = calloc( n, sizeof( int ) );
}
for ( size_t i = 0; i < n; i++ )
{
for ( size_t j = 0; j < n; j++ ) a[i][j] = i * n + j;
}
for ( size_t i = 0; i < n; i++ )
{
for ( size_t j = 0; j < n; j++ ) printf( "%2d ", a[i][j] );
putchar( '\n' );
}
for ( size_t i = 0; i < n; i++ )
{
free( a[i] );
}
free( a );
return 0;
}
The program output is the same as shown above.

Answer 2

numberOfConstPiArray[countKernel][col]
is getting an int* from numberOfConstPiArray[countKernel], then trying to dereference col'th element of this int*, and fails, as numberOfConstPiArray[countKernel] was not initialized with an reference to int array memory.
You may use instead:
int * numberOfConstPiArray = calloc(invariannumberOfUniqueKernels * kernelColumnCount, sizeof(int));
memset(numberOfConstPiArray, 0, invariannumberOfUniqueKernels * kernelColumnCount, sizeof(int));
...
numberOfConstPiArray[countKernel * kernelColumnCount + col] = 1;