/* 
   iobasic.c
   
   Run as: iobasic <n>

   What is does: allocates an array of n integegers, and traverses the
   array in sequential vs random order.

   Demonstrates the effect of the memory hierarchy and the limitations
   of the RAM model. 

   The RAM model assumes that all memory accesses take the same and
   complexity is modeled as 

   complexity = CPU complexity =  nb. CPU operations  

   When working with large data the cache and IO behaviour (nb. cache
   misses/page faults) are relevant and may actually be dominant.
   Theoretically this is modeled by

   cache complexity = nb blocks transferred between cache and main
   memory

   IO complexity = nb blocks transferred between main memory and disk  
 */

#include <stdio.h> 
#include <time.h> 
#include <assert.h> 
#include <stdlib.h> 

#define PRINT if(0)

int main(int argc, char** argv) {

  clock_t  t1, t2; 

  //need to run as: iobasic <n>  
  if (argc !=2)  {
    printf("usage: %s [n]\n", argv[0]);
    exit(1); 
  }

  //convert the argument to a long and print it 
  long n = atol(argv[1]); 
  printf("n=%ld (%.1f G) ints which require  %.1f GB RAM\n", 
	 n, 
	 (float)n/(1<<30), 
	 (float)n*sizeof(int)/(1<<30)); 


  //allocate  and initialize an array of n ints 
  t1 = clock(); 
  printf("%30s", "allocate and initialize a: "); 
  int* a = (int*)malloc(n * sizeof(int)); 
  assert(a); 
  for (long i=0; i<n; i++) {
    a[i]=i; 
  } 
  t2 = clock(); 
  printf("time elapsed %.3f seconds\n", (double)(t2-t1)/CLOCKS_PER_SEC); 
  

  //SEQUENTIAL ACCESS ARRAY A
  int sum = 0; 
  t1 = clock(); 
  printf("%30s", "sequential access: \n"); 
  for (long i=0; i<n; i++)  {
    sum =(sum + a[i])%2; 
    
    //print % progress
    PRINT {if (i% (n/100) ==0) printf("%d %% ", (int) (i/(n/100))); fflush(stdout);}
  }
  t2 = clock(); 
  printf("\n%30s time elapsed %.3f seconds\n", " ", (double)(t2-t1)/CLOCKS_PER_SEC); 



  //RANDOM ACCESS ARRAY A
  t1 = clock(); 
  printf("%30s", "random access: \n"); 
  for (long i=0; i<n; i++) {
    //generate a random index in 0..n
    long  k = random() % n; 
    sum =(sum + a[k])%2; 
  
    //print % progress
    PRINT  { if (i% (n/100) ==0) printf("%d %% ", (int) (i/(n/100)));fflush(stdout);}
  } 
  t2 = clock(); 
  printf("\n%30s time elapsed %.3f seconds\n"," ",  (double)(t2-t1)/CLOCKS_PER_SEC); 
  
  return 0; 
}