CSI 333 -- Fall 2010 -- Solutions to Midterm Exam
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Question I:
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(a) 36 in base 7 = 6 * 7^0 + 3 * 7^1 = 6 + 21 = 27 decimal. 

(b) In general, we cannot comment out C code using "/*" and "*/"
    since the code may itself contain comments that use  "/*" and "*/". 
    (Comments using "/*" and "*/" can't be nested.)

(c) In C, parameters are passed by value. Since we want the variables
    to be changed by the call to scanf (i.e., achieve the effect of
    call by reference), we need to pass pointers to variables.

(d) The assignment is not valid since the declaration makes only the    
    variable xx a pointer. (Variable yy is of type int; thus xx and
    yy are not of the same type.)

(e) The assignment is not valid since the type of variable x is
    struct employee. (x is NOT a pointer.) We can use the "->"
    construct only with pointer variables.

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Question II:
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Part (a):
---------

    210 in base 4 = 1 * 4^1 + 2 * 4^2 = 36 decimal.
    0.3 in base 4 = 3 * 4^(-1) = 3/4  = 0.75 decimal.

 So, 210.3 in base 4 = 36.75 decimal

Converting 36 (decimal) to binary: 

    36/2    quotient = 18    remainder = 0 
    18/2    quotient =  9    remainder = 0 
    9/2     quotient =  4    remainder = 1 
    4/2     quotient =  2    remainder = 0 
    2/2     quotient =  1    remainder = 0 
    1/2     quotient =  0    remainder = 1 

  So, 36 (decimal) = 100100 binary.

Converting 0.75 decimal to binary:

    0.75 x 2  =  1.5     bit = 1
    0.5 x 2   =  1.0     bit = 1
    
  So, 0.75 (decimal) = 0.11 binary.

So, 210.3 (base 4) = 100100.11 binary.
   
Part (b):
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   To convert +47 into binary:

      47/2    quotient = 23    remainder = 1 
      23/2    quotient = 11    remainder = 1 
      11/2    quotient =  5    remainder = 1 
      5/2     quotient =  2    remainder = 1 
      2/2     quotient =  1    remainder = 0 
      1/2     quotient =  0    remainder = 1 

   Thus, 47 (decimal) = 00101111  (8-bit binary)  

So, -47 (decimal) in 8-bit 2's complement form 
            = 11010001 (using the visual method)

Answer in hexadecimal form = D1.

Part (c):
---------

  When we convert a base 5 real number into decimal, each digit
to the right of the radix point is multiplied by an appropriate
positive power of the fraction 1/5. Since the fraction 1/5 = 0.2 has
a finite representation in decimal, all positive powers of 1/5 also
have finite representations.

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Question III:
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(a) The C program is given below.

    #include <stdio.h>
    int main(void) {
      printf("%d\n", sizeof(unsigned long));
      return 0;
    }

(b) Output:   

         8    3
         8   10


(c) Output:
     
     4
     a.out
     two
     three
     one

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Question IV:
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int  check_first_last (PNODE  head) {

  PNODE  tail;   /* To find the last node. */

  /* If the list is empty, the function must return -1. */
  if (head == NULL)
     return -1;

  /* The list is not empty. So, find the last node. */
  tail = head;

  /* The last node has its next field value NULL. */ 
  while (tail->next != NULL) {
     tail = tail->next;
  } /* End of while */

  /* Now, head and tail point to the first and the last nodes of 
     the list respectively. Compare the key values.
  */
  if (head->key == tail->key)
     return 1;  
  else 
     return 0;  

} /* End of check_first_last. */
       
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Question V:
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#include  <stdio.h>
#include  <stdlib.h>

int main(void) {

  char *input_file = "values.dat";   /* The name of input file.*/
  FILE *inf;    /* Input file pointer. */

  int  count; /* To find the number of integers in the file. */ 
  int  sum;   /* To find the sum of all the integers in the file. */ 
  int  max;   /* To find the maximum over all the integers in the file. */ 
  int  value; /* To read from input file. */

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

  /* Initialization. */
  count =  sum = 0;   
  max = 0;  /* This is ok since all the values in the input file are
               non-negative. */

  /* Loop through all the integer values in the input file to
     find the count, sum and max values.
  */

  while (fscanf(inf, "%d", &value) != EOF) {
     /* We have an integer value. Update count, sum and max.  */ 
     count++;  sum += value;
     if (value > max)
        max = value;
  } /* End of while. */

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

  /* Output values. (Since the file is guaranteed to have at least 
     one integer, there is no need to check whether count is 0.)
  */

  printf("Maximum value = %d\n", max);
  printf("Average value = %f\n", (float)sum/(float)count);

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

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