program stops responding when I run for loop in C - c

Question

Logic: I am trying to loop from 3 to 100 and put all prime numbers in that range in an array. I do this by first putting 2 manually into the array. After that, I loop over all the numbers and if a number is not divisible by all the prime numbers I have added to the array, it is a prime. The logic itself is not really valid in my opinion but this is a homework problem and I need to do it the way the professor wants it to be done.
So I try to loop and it gets to a point and the program just crashes. What am I doing wrong here?
int primeNums_lessThan100[25] = {2}; //25 is the size here because there are only 25 prime numbers under 100.
int primeNums_lessThan100_length = 1;
for(int counter = 3 ; counter < 100 ; counter++)
{
printf("Entered here!\n");
for(int array_idx = 0; array_idx < primeNums_lessThan100_length ; array_idx++)
{
if(counter % primeNums_lessThan100[array_idx] != 0)
{
printf("Entered here, TOO!\n");
primeNums_lessThan100[array_idx+1] = counter;
primeNums_lessThan100_length++;
}
else
{
continue;
}
}
}

Answer 1

You have a basic error in your logic. It's not enough to find out the counter value is coprime with some previously found prime to determine counter is prime itself. You need to test it is coprime with all primes found.
Suppose counter == 9 and you test it against the first item of your 'prime' numbers, which is 2. The result of 9 % 2 is of course 1, which is not equal zero and your program adds 9 to the array as a prime number. It happens even earlier for counter == 4 — first you find out it is divisible by 2 and reiterate the loop to find out in the next step 4 % 3 != 0 and add 4 to 'primes'.
As a further result the array overflows (you put more numbers into it than you expected and declared; actually, you put every natural number to it, because each N greater than 2 is not divisible by (N-1)), which means you eventually write past the end of the array, thus triggering an Undefined Behavior.
That displays also a basic error in the design: you do not check your array_idx against the array size, which might allow you to handle the error in some civilized manner.

Answer 2

the following proposed code:
cleanly compiles
performs the desired functionality
comments in code explain what is being done
note use of size_t for numbers that will never be less than 0
Note that all prior prime values are checked against the testPrime
Note that when a prime is found, it is placed in the next available entry in the array: `primeNums[]
and now, the proposed code:
#include <stdio.h> // puts(), printf()
// eliminate ''magic' numbers by giving them meaningful names
//25 is the size here because there are only 25 prime numbers under 100
#define MAX_PRIMES 25
#define TEST_LIMIT 100
int main( void )
{
// declare prime number array and init to 0
int primeNums[ MAX_PRIMES ] = {0};
// pre-set first prime number
primeNums[0] = 2;
// init number of primes found
size_t primeCount = 1;
// test all values of interest
for( int testPrime = 3 ; testPrime < TEST_LIMIT ; testPrime++ )
{
// init 'prime' indicator, 1 means value is prime
size_t Prime = 1;
for( size_t prime_idx = 0; prime_idx < MAX_PRIMES; prime_idx++ )
{
// check that array of prime values contains a valid prime
if( primeNums[ prime_idx ] )
{
if( testPrime % primeNums[ prime_idx ] == 0 )
{
// indicate not a prime number
Prime = 0;
}
}
else
{ // else, all previously found prime numbers checked
break;
} // end if
} // end for each prime already found
if( Prime )
{ // then test number was prime
// update the array of primes
primeNums[ primeCount ] = testPrime;
// update the count of primes found
primeCount++;
}
}
// display the found primes
for( size_t i = 0; i < primeCount; i++ )
{
printf( "%d, ", primeNums[i] );
}
// assure the data gets displayed on the terminal before the program exits
puts( "" );
}
a run of the proposed code results in:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,