CS 2734 Computer Organization II -- Spring 2001
Laboratory 2 [Jan 17, 19]: Binary Arithmetic

The basic laboratory:
Laboratory 2 is an introduction to binary arithmetic and representations of signed numbers.

For this laboratory, you are to experiment with the bit pattern used to represent an int in computers that use 2's complement to represent 32-bit integers. (See text, Section 4.2, pages 210-220. In C/C++ on Sun hardware the int type is a 32-bit 2's complement signed integer.)

Given either an "ordinary" integer (in decimal) or a bit pattern, it is easy to get the other representation, say, with a simple program like the following: 

#include <stdio.h>
#include <ctype.h>
void main(void){
   int i;
   char ch;
   for( ; ; ) {
      while (isspace(ch = getchar()))
         ;
      if (ch == 'x') {
         scanf("%x", &i);
         printf("  Decimal: %10i\n", i);
         printf("  Bits: %08x \n", i);
      }
      else if (ch == 'i') {
         scanf("%i", &i);
         printf("  Decimal: %10i\n", i);
         printf("  Bits: %08x \n", i);
      }
      else break;
   }
}
Typical output looks like the following (user input in boldface): 
i 47
  Decimal:         47
  Bits: 0000002f 
i -3
  Decimal:         -3
  Bits: fffffffd 
x ffffffff
  Decimal:         -1
  Bits: ffffffff 
q
For this laboratory, experiment with the program by entering various integers, including 0, 1, 2, -2, -3, 2147483647, 2147483648 (overflow), -2147483648, and -2147483649 (underflow).
Sample Exam Questions:
Here are some sample problems of the type you should expect on the first quiz and the first exam.

  1. Convert 189 to:
    a) An unsigned binary number
    b) A 16-bit two's complement number

  2. Convert -189 to:
    a) A 16-bit two's complement number

  3. Convert binary number 10011111001 to:
    a) octal
    b) hexadecimal
    c) decimal

  4. Perform the following binary addition:
    10000111 + 10111011

  5. Convert the hexadecimal number AB32 to:
    a) binary
    b) octal
    c) decimal

The answers are:
  1. a) 10111101
    b) 0000000010111101
  2. b) 1111111101000011
  3. a) 2371
    b) 4F9
    c) 1273
  4. 101000010
  5. a) 1010 1011 0011 0010
    b) 125462
    c) 43,826

Extra for those bored with the above:
Here is a more interesting program that converts between number bases without using the C printf formatting: 
/* writebase.c: convert between number bases in C
 *   This program does not handle bases > 10 nor
 *   does it handle negative numbers.
 *   Written by Neal R. Wagner, 7 Sept 1999
 */
#include <stdio.h>
#include <ctype.h>

void dumpout(int n, int base);
void writebase(int n, int base);
int readbase(int base);

void main(void)
{
   int n, base;
   for( ; ; ) {
      printf("input base (in base 10)>>>");
      base = readbase(10);
      if (base <= 0) break;
      printf("Input number (in base ");
      writebase(base, 10);
      printf(")>>>");
      n = readbase(base);
      dumpout(n, 10);
      dumpout(n, 8);
      dumpout(n, 2);
      printf("--------------------------\n");
   }
}

void dumpout(int n, int base)
{
   printf("Base "); writebase(base, 10);
   printf(": ");
   writebase(n, base); putchar('\n');
}

void writebase(int n, int base)
{
   if (n != 0) {
      writebase(n/base, base);
      putchar((n%base) + '0');
   }
}

int readbase(int base)
{
   char c;
   int i = 0;
   while (isdigit(c = getchar()))
      i = base*i + (c - '0');
   return i;
}
Here is some sample output:
four06% writebase
input base (in base 10)>>>8
Input number (in base 8)>>>23456
Base 10: 10030
Base 8: 23456
Base 2: 10011100101110
--------------------------
input base (in base 10)>>>10
Input number (in base 10)>>>99999
Base 10: 99999
Base 8: 303237
Base 2: 11000011010011111
--------------------------
input base (in base 10)>>>0

Revision date: 2001-01-19. (Please use ISO 8601, the International Standard.)