Infinite while loop issue with a conditional operation function

Question

My code gets stuck in an infinite loop. The current number is halved, When the next number is even, the function should execute 2n+1. If odd, it should execute 3n + 1. Once either operation is executed it should halve it again and loop until n = 1. Here is the code:

#include "stdio.h"
#include "assert.h" // ?

long int hailstone(long int k);

int main(void) {
  long int n = 77;
  hailstone(n);
  
  return 0;
}

long int hailstone(long int k) {
  while (k != 1) {
    k = k/2;
    if (k % 2 == 0) {
          k = 2 * k + 1;
          printf("%lu", k);
    
    } else if (k % 2 != 0) {
          k = 3 * k + 1;
          printf("%lu", k);
      
    } else if (k == 1) {
          printf("blue sky!");
    }
  }
}

Would a particular assertion help the compiler execute the code as expected?

Answer 1

As I understand your code

If the value of n is pair ,you make this n/2 , if not you display 3*n+1

My code :

output : 116,58,29,44,22,11,17,26,13,20,10,5,8,4,2,1

#include "stdio.h"
#include "assert.h" // ?


long int hailstone(long int k);

int main(void)
{
    long int n = 77;
    hailstone(n);
    return 0;
}

long int hailstone(long int k)
{
    while (k != 1)
    {
       if ( k % 2 ==0)
       {
           printf("k = %lu\n",k/2);
           k/=2;   // k=k/2
       }
       else
       {
           k = 3 * k + 1;
       }
    }
        printf("blue sky!\n");
    
}