'How to count this operation for (int interval = n/2; interval > 0; interval /= 2) using counting primitive operation?

I was confused how to label this for (int interval = n/2; interval > 0; interval /= 2) with counting operation and estimating this operation so that I can get that exact Running time of this shell short algorithm

for (int interval = n/2; interval > 0; interval /= 2)  
    {  
        for (int i = interval; i < n; i += 1)  
        {  
            /* store a[i] to the variable temp and make the ith position empty */  
            int temp = a[i];  
            int j;        
            for (j = i; j >= interval && a[j - interval] > temp; j -= interval)  
                a[j] = a[j - interval];  
              
            // put temp (the original a[i]) in its correct position  
            a[j] = temp;  
        }  
    }  

               1    (1 or log base 2)?
int interval   =     n / 2

now here with condition that I am referring as i < 0 which is n+1 in other operation sample.

       (n+1)
interval > 0

I know that the operation of this is log base2

         log base 2  
interval /= 2
performancebig-ocomplexity-theory


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