The document discusses integer representation and arithmetic logic unit (ALU) operations in computers. It describes how integers are represented using binary digits and two's complement notation. The ALU performs calculations on these binary integers. It can handle standard integer arithmetic as well as floating point numbers, potentially with a separate floating point unit.

1. DATA REPRESENTATION

2. ARITHMETIC & LOGIC UNIT
 Does the calculations
 Everything else in the computer is there to
service this unit
 Handles integers
 May handle floating point numbers
 May be separate FPU

3. ALU INPUTS AND OUTPUTS

4. INTEGER REPRESENTATION
 Only have binary digits (0 & 1) to represent
numbers.
 Positive numbers stored in binary
 e.g. 41=00101001
 Sign-Magnitude
 Two’s compliment

5. An 8 -bit word can represent the numbers from
0 to 255,including
00000000 = 0
00000001 = 1
00101001 = 41
10000000 = 128
11111111 = 255
If an n-bit sequence of binary digits an-1an-2…a1a0 is
interpreted as an unsigned integer A, its value is
A= i
n
i
i
a



1
0
2

6. SIGN-MAGNITUDE
 Left most bit is sign bit
 0 means positive
 1 means negative
 If there are n bits, then rightmost n-1 bits hold the magnitude of
the integer.
 +18 = 00010010
 -18 = 10010010

7.  Problems
 Need to consider both sign and magnitude in
arithmetic
 Two representations of zero (+0 and -0)
(e.g) +010 =00000000
-010 =10000000

8. CHARACTERISTICS OF TWOS COMPLEMENT
REPRESENTATION AND ARITHMETIC

9. BENEFITS
 One representation of zero
 Arithmetic works easily
 Negating is fairly easy
 3 = 00000011
 Boolean complement gives 11111100
 Add 1 to LSB gives 11111101

10. VALUE BOX FOR CONVERSION BETWEEN
TWOS COMPLEMENT BINARY AND DECIMAL

12. SOLUTION
Move the sign bit to the new leftmost position and fill in with
copies of the sign bit.(Sign Extension)
-18 = 11101110(twos complement,8 bits)
-18 = 1111111111101110(twos complement,16 bits)

13. NEGATION
Rules:
1) Take the Boolean complement of each bit of the
integer.(including the sign bit)
2) Treating the result as an unsigned binary integer , add
1.
Eg) +18 = 00010010
bitwise complement = 11101101
1 +
11101110 = -18

14. NEGATION SPECIAL CASE 1
 0 = 00000000
 Bitwise not 11111111
 Add 1 to LSB +1
 Result 1 00000000
 Overflow is ignored, so:
 - 0 = 0

15. NEGATION SPECIAL CASE 2
 -128 = 10000000
 bitwise not 01111111
 Add 1 to LSB +1
 Result 10000000
 So:
 -(-128) = -128