C programming: Replacing if statements - c

Question

Hello a beginner here who needs some of your help. My C program is good and does what it is supposed to do only that it is not supposed to use any kind of if statements. I wrote it that way as I saw it would be easier so that I can then replace the if statements. I have been trying to replace the if statements but am now stuck. What can I use instead of the if statement to still produce the same output.
The program is supposed to generate a sequence of thirty random integers between 0 and 9 and then print out the sequence both forward and backwards. Then print out a count of how many times each number between 0 and 9 appeared in the sequence.
This is the output
Here is a sequence of 30 random numbers between 0 and 9:
3 6 7 5 3 5 6 2 9 1 2 7 0 9 3 6 0 6 2 6 1 8 7 9 2 0 2 3 7 5
Printing them backwards, that's:
5 7 3 2 0 2 9 7 8 1 6 2 6 0 6 3 9 0 7 2 1 9 2 6 5 3 5 7 6 3
There were 3 0's
There were 2 1's
There were 5 2's
There were 4 3's
There were no 4's
There were 3 5's
There were 5 6's
There were 4 7's
There was only 1 8
There were 3 9's
This is my C program
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i, j, array[30]={0}, count=0,check;
srand(time(NULL));
for(i=0;i<30;i++)
array[i]=rand()%10;
for(i=0;i<30;i++)
printf("%d ",array[i]);
printf("\n\n");
for(i=29;i>=0;i--)
printf("%d ",array[i]);
printf("\n\n");
for(i=0;i<30;i++){
check=array[i];
if(array[i]!=-1)
array[i]=-1;
if(check == -1)
continue;
count =1;
for(j=0;j<30;j++){
if((i==j) || (array[j]==-1))
continue;
if(check==array[j]){
count++;
array[j]=-1;
}
}
printf("There were %d %d's\n",count,check);
}
return 0;
}

Answer 1

You'll understand the algorithm from comments:
#include <stdio.h>
#include <stdlib.h>
//time.h is needed for time()
#include <time.h>
int main()
{
int i, array[30] = {0};
srand(time(NULL));
//generate and print 30 random numbers
for(i = 0; i < 30; i++){
array[i] = rand() % 10;
printf("%d ", array[i]);
}
puts("\n\n");
//print these numbers backwards
for(i = 29; i >= 0; i--)
printf("%d ",array[i]);
puts("\n\n");
// print out a count of how many times each number
// between 0 and 9 appeared in the sequence.
int count[10] = {0};
for(i = 0; i < 30; i++)
count[array[i]]++;
//output the count for each number
for(i = 0; i < 10; i++)
printf("There were %d %d's\n",count[i], i);
return 0;
}
Output:
9 2 3 9 8 4 3 8 1 3 6 4 3 2 5 3 2 3 0 1 9 0 3 5 1 3 3 8 2 0
0 2 8 3 3 1 5 3 0 9 1 0 3 2 3 5 2 3 4 6 3 1 8 3 4 8 9 3 2 9
There were 3 0's
There were 3 1's
There were 4 2's
There were 9 3's
There were 2 4's
There were 2 5's
There were 1 6's
There were 0 7's
There were 3 8's
There were 3 9's

Answer 2

The following is a simplification of your original source while removing the if statements. There are implied if statements in several places where a logical expression is used as part of a source code statement.
For instance for(j = 0; j < 30 && match >= 0; j++) has several logical expressions but no if appears in this statement. The logical expressions are j < 30 and match >= 0 and the complete expression of j < 30 && match >= 0.
This example uses a logical expression and the evaluation short circuit behavior of the C compiler (see Short-circuit evaluation in Wikipedia) in the statement array[j] == match && ++count && (array[j] = -1); so that if the logical expression array[j] == match evaluates to false then the rest of the statement will not be executed.
We also depend on the preincrement operator with the ++count to increment count and then take the resulting value to check if it is false (zero) or true (non-zero). Since the variable count is initialized to zero and when incremented will always be non-zero then the next logical expression in the statement is evaluated the (array[j] = -1). We put the assignment statement within parenthesis to enforce the order of evaluation. We want the variable array[j] to be assigned the value of -1 and for the result to then be used in the logical statement. Since this is the last logical expression of the entire logical statement, whether it evaluates to false (zero) or true (non-zero) doesn't matter as what we want is the side effect of assigning the value of -1 to the array element.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
int i, array[30] = {0};
srand(time(NULL));
for(i = 0; i < 30; i++)
array[i] = rand() % 10;
for(i = 0; i < 30; i++)
printf("%d ", array[i]);
printf("\n\n");
for(i = 29; i >= 0; i--)
printf("%d ",array[i]);
printf("\n\n");
for(i = 0; i < 30; i++){
int j;
int count = 0;
int match = array[i];
for(j = 0; j < 30 && match >= 0; j++){
array[j] == match && ++count && (array[j] = -1);
// replaces the following if as value of count is
// tested after it is incremented so will always be nonzero.
// if (array[j] == match) {
// count++; array[j] = -1;
// }
}
// if this is a valid array element value we are trying to match
// then print the count and the value being matched. printf()
// is a function that returns an int indicating number of character written.
match >= 0 && printf("There were %d %d's\n", count, match);
}
return 0;
}
An example output.
8 2 4 0 8 0 8 1 1 4 6 9 3 9 7 6 3 9 0 1 0 7 1 2 4 0 3 0 2 3
3 2 0 3 0 4 2 1 7 0 1 0 9 3 6 7 9 3 9 6 4 1 1 8 0 8 0 4 2 8
There were 3 8's
There were 3 2's
There were 3 4's
There were 6 0's
There were 4 1's
There were 2 6's
There were 3 9's
There were 4 3's
There were 2 7's