Write a C program that defines a 2-dimentional integer array
called A [50][50]. Then the elements of this array should
randomly be initialized either to 1 or 0. The program should
then print out all the elements in the diagonal (i.e.
a[0][0], a[1][1],a[2][2], a[3][3], ……..a[49][49]). Finally,
print out how many zeros and ones in the diagonal.
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Answer Posted / cfuzz
/* this is my code, only thing I'm missing is counting
zeros...WHAT THE HELL am I doing WRONG?*/
#include <stdio.h>
#define ROWS 50
#define COLS 50
int count_occur(int A[], int num_elements, int value);
int main(void)
{
int A[ROWS][COLS];
int i=0, j=0;
int num_occ, value=0;
/* Initializing*/
for(i=0; i < ROWS; i++) {
for(j=0; j < COLS; j++) {
A[i][j] = 0;
A[i][j] = rand() % 2;
}
}
for(i=0; i < ROWS; i++) {
for(j=0; j < COLS; j++) {
if (i == j){
printf("%2d", A[i][j]);
}
}
}
for(value=0; value<1; value++)
{
num_occ = count_occur(A, 50, value);
if (value = 1){
printf("\n\nThe value %d was found %d times.\n", value,
num_occ);
}
else if (value = 0){
printf("\n\nThe value %d was found %d times.\n", value,
num_occ);
}
}
}
int count_occur(int A[], int num_elements, int value)
/* checks array a for number of occurrances of value */
{
int i, count=0;
for (i=0; i<num_elements; i++)
{
if (A[i] == value)
{
++count; /* it was found */
}
}
return(count);
}
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