CareerCup

1. Allocate some memory and keep pushing the numbers as it comes.
2. If u run out of memory, realloc
3. when u get 500, return the pointer.

Am I missing something?

Define ArrayList<Integer> alist=new ArrayList<Integer>(); Check each time, when the size of alist is 500 more than the former size, then return alist.toArray(), is it right?

If I am getting the problem correctly, then it means whenever 500 appears we have to print all the elements before it.
Hence we can keep on storing numbers in a file and whenever a 500 comes, we should print all the content of file on stdout and make the file empty.

while(n<count){printf ("%d ",numbers[n]);/* n is always less than count.*/
n++;}
} /*loops forever*/

Hello Anonymous. Thank you very much for posting your code. I have been stuck for a couple of weeks on Ex7.1, Pg 330 of Beginning C, Ivor Horton. I have adapted your code to enter an unspecified number of float values, and then print them out with the average as well. You have saved me much time.
Best regards.

/*realloc_for_averaging_value_of_floats_fri14Sept2012_16:30  */


#include <stdio.h>
#include <stdlib.h>
#define TRUE 1

int main (int argc, char ** argv[])
{
 float input =0;
 int count=0 ,n =0;
 float * numbers = NULL;
 float * more_numbers;
 float sum = 0.0;

while(TRUE)
{
do {
    printf ("Enter an floating point value (0 to end): ");
    scanf ("%f", &input);
    count++;

more_numbers = (float*) realloc (numbers, count * sizeof(float));

if (more_numbers!=NULL) 
{
   numbers=more_numbers;
   numbers[count-1]=input;
}
else 
  {
    free (numbers);
    puts ("Error (re)allocating memory");
    exit (TRUE);
  }
} while (input!=0);

  printf("Numbers entered: ");
    while(n<count){printf("%f ",numbers[n]);  /* n is always less than count.*/
    n++;}                                                        
                          /*need n++ otherwise loops forever*/
 n=0;
 while(n<count)
   {
     sum += numbers[n];      /*Add numbers together*/ 
     n++;
   };
printf("\n Average of floats = %f \n",sum/(count-1));/* Divide sum / count = average.*/
 }
   return 0;

 }


/* Success Fri Sept 14 13:29 . That was hard work.*/
/* Always looks simple when working.*/
/* Next step is to use a function to work out the average.*/
/*Anonymous on July 04, 2012*/