CSI 333 -- Fall 2011 -- Solutions to Midterm Exam
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Question I:
-----------

Part (a):
---------

Converting 259 (decimal) to octal: 

    259/8   quotient = 32    remainder = 3
    32/8    quotient = 4     remainder = 0
    4/8     quotient = 0     remainder = 4

  So, 259 (decimal) = 403 (octal)


Part (b):
---------

    123 in base 5 = 3 * 5^0 + 2 * 5^1 + 1 * 5^2  = 38 decimal.

    Converting 38 (decimal) to hexadecimal: 

    38/16   quotient = 2    remainder = 6
    2/16    quotient = 0    remainder = 2

  So, 123 (base 5) = 26 (hex)

Part (b):
---------

Converting 70 (decimal) to binary: 

    70/2    quotient = 35    remainder = 0 
    35/2    quotient = 17    remainder = 1 
    17/2    quotient =  8    remainder = 1 
    8/2     quotient =  4    remainder = 0 
    4/2     quotient =  2    remainder = 0 
    2/2     quotient =  1    remainder = 0 
    1/2     quotient =  0    remainder = 1 

  So, 70 (decimal) = 1000110 binary.

Converting 0.15 decimal to binary:

    0.15 x 2  =  0.3     bit = 0
    0.3 x 2   =  0.6     bit = 0

    0.6 x 2   =  1.2     bit = 1
    0.2 x 2   =  0.4     bit = 0
    0.4 x 2   =  0.8     bit = 0
    0.8 x 2   =  1.6     bit = 1

  Now, the bit pattern 1001 repeats indefinitely.
    
                           ----
  So, 0.15 (decimal) = 0.001001 binary.

                                ----
So, 70.15 (decimal) = 1000110.001001 binary.
   
Part (d):
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   To convert +29 into binary:

      29/2    quotient = 14    remainder = 1 
      14/2    quotient =  7    remainder = 0 
      7/2     quotient =  3    remainder = 1 
      3/2     quotient =  1    remainder = 1 
      1/2     quotient =  0    remainder = 1 

   Thus, 29 (decimal) = 11101  (binary)  
                      = 0001 1101  (8-bit binary)  

So, -29 (decimal) in 8-bit 2's complement form 
            = 1110 0011 (using the visual method)

Answer in hexadecimal form = E3.

Part (e):
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#define  DIVISOR   10

int  count_digits (int x) {

  /* Returns the number of digits in the decimal integer x.
     It is assumed that x is positive.

     Idea: Successive integer divisions of x by 10. Each division
           eliminates one digit of x.
  */

  int  count = 0;  /* The number of digits in x -- to be computed. */

  while (x > 0) {
    count++; x /= DIVISOR;
  }

  return count;
} /* End of count_digits. */


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Question II:
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(a) Output:

      r = 5
      y = 5
      y = 6
      r = 2

(b) Output:   

      21   21   31
     -14  -14  -14

(c) Output for Part (i):

      4
      6

    Output for Part (i):

      2
      5

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Question III:
-------------

int  check_list (PNODE  head) {

   PNODE  prev;  /* Temporary. */ 

  /* If the list is empty or has only one node, the function */
  /* must return 0.                                          */

  if ((head == NULL) || (head->next = NULL))
     return 0;

  /* The list has two or more nodes. Traverse the list and */
  /* check if there are two successive nodes with negative */
  /* key values.                                           */

  prev = head;  head = head->next;

  while (head != NULL) {
     if ((prev->key < 0) && (head->key < 0)) 
        /* Found two successive nodes with negative key values. */
        return 1;   
     else {
        prev = head;  head = head->next;
     }
  } /* End of while */

  /* If we reach this point, then the list is not special. */
  return 0;  

} /* End of check_list. */
       
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Question V:
-----------

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

#define  DIVISOR   17

int main(void) {

  char *infname  = "infile.dat";   /* The name of input file.*/
  char *outfname = "outfile.dat";   /* The name of output file.*/

  FILE *infp, *outfp;    /* Input and output file pointers. */

  int  value; /* To read from input file. */

  /* Open the input file for reading. */
  if ((infp = fopen(infname, "r")) == NULL) {
     fprintf(stderr, "Couldn't open input file -- %s\n", infname);
     exit(1);
  }

  /* Open the output file for writing. */
  if ((outfp = fopen(outfname, "w")) == NULL) {
     fprintf(stderr, "Couldn't open output file -- %s\n", outfname);
     exit(1);
  }

  /* Loop through all the integer values in the input file and
     write appropriate values to the output file.
  */

  while (fscanf(infp, "%d", &value) != EOF) {

     /* We have an integer value. If it is negative or a multiple
        of 17, then write to the output file.  
     */ 
     if ((value < 0) || ((value % DIVISOR) == 0)) 
        fprintf(outfp, "%d\n", value);

  } /* End of while. */

  /* Close the input file. */
  if (fclose(infp) == EOF) { /* Error in closing input file */
     fprintf(stderr, "Error in closing file %s\n", infname);
     exit(1);
  }

  /* Close the output file. */
  if (fclose(outfp) == EOF) { /* Error in closing output file */
     fprintf(stderr, "Error in closing file %s\n", outfname);
     exit(1);
  }

  return 0;
} /* End of main. */

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