How to access elements of a matrix array via pointers in C? - c

Question

I can't seem to find any info on how to access elements of an array via pointer in a function, I tried following multiple answers but none of them seem to work for me.
My task is next: Write a program in C with m x n dimension with elements being randomly generated from 0 to 9. Using two new functions calculate the sum of even elements and count the number of elements being equal to zero.
#include <stdio.h>
#include <stdlib.h>
void SumEven(int *a, int n, int m, int *sum){
}
void EqualToZero(int *a, int n, int m, int *number){
}
int main()
{
int** a;
int m, n, l, i, j, r, sum;
printf("Enter number of columns for matrix: ");
scanf("%d", &m);
printf("Enter number of rows for matrix: ");
scanf("%d", &n);
a = (int **) malloc(m*sizeof(int));
for (l = 0 ; l < m ; l++){
a[l] = (int **) malloc(n*sizeof(int));
}
time_t t;
srand((unsigned)time(&t));
printf("\n");
printf("Your matrix is:\n");
printf("\n");
for(i = 0 ; i < m ; i++){
for(j = 0 ; j < n ; j++){
r = rand() % 10;
a[i][j] = r;
printf("%d ", r);
}
printf("\n");
}
printf("\n");
SumEven(&a, n, m);
return(0);
}
As you can see in the provided code I left those functions empty as I don't know how to pass the matrix to them and access their elements so I can be able to print my results.
Also my pseudo code for the logic for the functions themselves are:
if(a[i][j] % 2 == 0)
printf("%d ", a[i][j])
and
if(a[i][j] == 0)
printf("%d ", a[i][j])
Also those parameters of the function are predefined in my task, so I have to follow them.
EDIT: I also don't know if I'm even passing the same matrix to the function with SumEven(&a, n, m);. I tried outputing the address of the matrix and using printf("%d", &a) to display an address both from main() and SumEven() functions.

Answer 1

This code may help. It does the following:
1. For an arbitrary array of integers, sum the elements of the array
- using a pointer to the SUM function
2. For an arbitrary array of integers, count the number of zero elements in
the array - using a pointer to the COUNTZERO function
#include <stdio.h>
#include <stdlib.h>
// sum the elements of the matrix
void sum(int* arr, int rows, int cols, int* result)
{
int sum = 0;
int i, j;
for (i = 0; i < rows; i++) {
for (j = 0; j < cols; j++) {
sum = sum + arr[i*cols + j];
}
}
*result = sum;
}
// count the number of zero elements in the matrix
void countZero(int* arr, int rows, int cols, int* result)
{
int count = 0;
int i, j;
for (i = 0; i < rows; i++) {
for (j = 0; j < cols; j++) {
if (arr[i*cols + j] ==0) count = count + 1;
}
}
*result = count;
}
// arbitrary initialisation of 2D array of ints (force last entry of the array to equal zero - for testing purposes)
void init2D(int *arr, int rows, int cols) {
int i, j;
for (i = 0; i < rows; i++) {
for (j = 0; j < cols; j++) {
arr[i*cols + j] = 1;
}
}
// use this to test the countZero function
arr[(rows-1)*(cols-1)] = 0;
}
int main() {
int *array; // will hold a 2D array of integers
int N = 10; // arbitrary number of rows
int M = 5; // arbitrary num cols
// 2D array of integers expressed as one "contiguous row" of memory
// make sure your indexing is correct when referenceing the array for (i,j)th element
array = (int*)malloc(sizeof(int)*N*M);
if (array != NULL) {
init2D(array, N, M);
}
// the function pointer
void(*general)(int*,int,int,int*);
// will contain the sum result
int sumAll = 0;
int* ptsumAll = &sumAll;
// make the function pointer point to the sum function
general = ∑
// sum the contents of the array
general(array,N,M, ptsumAll);
printf("sum of array elements: %d\n", *ptsumAll);
// contains a count of the zero elements in the array
int count =0;
int* ptcount = &count;
// make the function pointer point to the count function
general = &countZero;
// count the number of zero element in the array
general(array, N, M,ptcount);
printf("number of zeros: %d\n", *ptcount);
free(array);
return 0;
}
some references:
https://www.cs.swarthmore.edu/~newhall/unixhelp/C_arrays.html
http://www.cprogramming.com/tutorial/function-pointers.html

Answer 2

I have added comments to help you with the code.
#include <stdio.h>
#include <stdlib.h>
void SumEven(int *a, int n, int m, int *sum){
//put this statement in 2 nested for loops of size n and m
if(a[i][j] % 2 == 0)
sum += a[i][j];
}
void EqualToZero(int *a, int n, int m, int *number){
//put this statement in 2 nested for loops of size n and m
if(a[i][j] == 0)
number++;
}
int main()
{
int** a;
int m, n, l, i, j, r, sum;
printf("Enter number of columns for matrix: ");
scanf("%d", &m);
printf("Enter number of rows for matrix: ");
scanf("%d", &n);
a = (int **) malloc(m*sizeof(int));
//should be m*sizeof(int*)
for (l = 0 ; l < m ; l++){
a[l] = (int **) malloc(n*sizeof(int));
//should be cast as (int*)
}
//I suggest you take look at declaring 2d arrays in C
time_t t;
srand((unsigned)time(&t));
printf("\n");
printf("Your matrix is:\n");
printf("\n");
for(i = 0 ; i < m ; i++){
for(j = 0 ; j < n ; j++){
r = rand() % 10;
a[i][j] = r;
printf("%d ", r);
}
printf("\n");
}
printf("\n");
SumEven(&a, n, m);
//need to pass &sum to this function. Also make sure it is initialized to 0
//call EqualToZero() function with proper parameters.
return(0);
//return 0; not return(0);
}

Answer 3

These will be your function prototypes:
void SumEven(int **a, int n, int m,int *sum);
void EqualToZero(int **a, int n, int m,int *number);
since you are passing a(double pointer) from calling then the there should be a double pointer(int **a) to receive it.
Calling:
SumEven(a, n, m,&sum);
EqualToZero(a, n, m,&number);
And this is how you can access the array inside your function:
void SumEven(int **a, int n, int m,int *sum){
int i,j,tsum=0;
for(i = 0 ; i < m ; i++){
for(j = 0 ; j < n ; j++){
if(a[i][j] % 2 == 0)
{
tsum+=a[i][j];
printf("%d ",a[i][j]);
}
}
}
*sum=tsum;
}
Also there is an error in this line a[l] = (int **) malloc(n*sizeof(int)); (‘int**’ to ‘int*’assignment),it should be a[l] = (int *) malloc(n*sizeof(int));.

Answer 4

Here's an example, given a 3D array
int buffer[5][7][6];
An element at location [2][1][2] can be accessed as buffer[2][1][2] or *( *( *(buffer + 2) + 1) + 2).
Reference

Answer 5

if(( *(*(a + i) + j) % 2 ) == 0 )
printf("%d", *(*(a + i) + j) )
if( *(*(a + i) + j) == 0 )
printf("%d", *(*(a + i) + j) )
This is how you do it.