Chapter 2_summary

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Transcript Chapter 2_summary

Chapter 2
Summary
Overview
• Numbers
– Decimal, Binary, Octal, Hexadecimal
– Their relationship
•
•
•
•
Sign-magnitude
One’s complement
Two’s complement
Arithmetic operation
Positive Numbers
• A computer represents positive integers in
binary
• Three methods for representing negative
numbers (signed numbers) are
– Sign-magnitude
– One’s complement
– Two’s complement
Sign Magnitude
• Representation the number’s sign and
magnitude (value)
• Positive numbers  0 and negative
numbers  1
• e.g. 01000000 = 64 and 11000000 = -64
• Easy for human to understand but it requires
some special logic for arithmetic operations
(addition, subtraction)
Arithmetic Operation
• Computers mostly storing information ,
particularly integers, in byte form
• In this section we look how addition and
subtraction are performed in computers
• To perform mathematical functions, a CPU
contains a component called ALU (arithmetic
logic unit) – to perform arithmetic operations
(addition & subtraction) and logic operation (in
next chapter)
One’s Complement
• The negative number is represented by flipping
the number’s bits
• E.g., 01001001 becomes 10110110
• E.g. consider 103 + -97
01100111 + (-01100001) = 01100111 + 10011110

= 100000101 (9 bits)
 00000101 + 1 = 00000110 (8 bits)
 103 – 97 = 6 = 00000110
 Try 113 + -42 = ?? and –75 + 13 = ??
Two’s Complement
• Makes addition and subtraction simple
• The fact that all numbers are stored in 2’s
complement form allows the ALU to use
just one circuit for all basic mathematical
operations
• Visual Basic uses 2’s complement
representations of numbers
Two’s Complement
• Rules
–
–
–
–
Just add all the bits
Throw away EAC (“end around carry”)
if a – b becomes a + (-b)
e.g.
111111 (-1)
10110 (-10)
+ 001000
(8)
+ 11101 (-3)
1000111
(7)
110011 (-13)
• try again –75 + 13 in 2’s complement
Assignment 1
• Question 1.9
(i) -117 – 145
- 117 = 10001011 (2’s)
- 145 = 01101111 (2’s)
 Answer : 11111010
Verify:
 00000110 ≠ - 262
(ii) –90 – 133
-90 = 10100110 (2’s)
-133 = 01111011 (2’s)
 Answer: 100100001
Verify:
 11011111 = -223