Transcript CSC 150 UNGRADED QUIZ - Concordia University Wisconsin

PROCESSOR UNIT I
(1) Job of the processor unit.
(2) How can it do this?
(3) Binary translation schemes
[base 2, ASCII, machine code].
1. Job of the processor unit.
 A. Review the diagram of the processor
unit’s CPU and memory.
 B. The job of the CPU is to fetch and
execute program instructions from RAM.
 The control unit does the fetch and execute.
 The ALU does arithmetic and logic.
 As the instructions are executed, data is
turned into information.
2. How can the processor unit
do this?
 A. In order to run a program:
 All program instructions and data must be
loaded into RAM.
 BUT
 1) The CPU only “understands” binary
 Therefore?
 2) All data and instructions must be in
binary form.
3. Binary translation schemes.
 There are 3 main things which need to be
translated:
 1) NUMBERS
 2) CHARACTERS
 3) PROGRAM INSTRUCTIONS.
 Each has its own binary translation scheme.
Binary schemes.
 ITEM
 1) NUMBERS
 2) CHARACTERS
 3) INSTRUCTIONS
 TRANSLATION
 SCHEME
 1) BASE 2 BINARY
FORM.
 2) ASCII BINARY
FORM
 3) MACHINE
LANGUAGE
BINARY FORM
3.1 Base 2 binary form.




Numbers are represented in Base 2.
What is the idea of base 2?
Compare base 10.
Base 10:




1) 10 digits (0-9)
2) Place values are powers of 10
e.g. 3
4
5
100 10 1 <---- place values.
Base 2 binary form continued.
 Base 2:




1) 2 digits (0-1)
2) Place values are powers of 2
e.g. 1
0 1 1 0 1
32 16
8
4
2
1 <--- place values.
Base 2 binary form continued.
 Example:
 What is the following binary number
equivalent to in base 10?
1 0 1 0 1 1
0
Base 2 binary form continued.







Recall 1 byte = 8 bits.
Largest number in 1 byte
1111 1111 = 255
[can store 256 numbers 0-255].
In 2 bytes?
1111 1111 1111 1111
Hint: the 16th place value = 32,768.
Base 2 binary form continued.
 Counting in binary.

0

1

10

11

100

101
3.2. ASCII binary form.
 For characters, a coding scheme is used,
such as as ASCII.
 Character
Code.
Binary
 ‘0’
48
110000
 ‘9’
57
111001
 ‘A’
65
1000001
 ‘Z’
90
1100001
ASCII binary form continued.
 Every character has a code, whether
printing or not.

Character
Code Binary
 E.g. “Return”
10
1010

“Line feed” 13
1101

“Escape”
27
11011
ASCII binary form continued.
 But now there is a problem. Binary strings
are now ambiguous. E.g. What is

100001
 A number?
Or
A character
 65 in Base 2
ASCII 65 = “A”
 How does the H/W know?
ASCII binary form continued.
 By itself, the H/W does not know!
 The S/W has to tell the H/W how the data is
to be interpreted. E.G. in C++,
 descriptors are used to specify the data type.
 char Name[20]
float Amount;
 [character string]
[number]
3.3 Instructions.
 A. Each instruction in the S/W is translated
into 1 or more binary instructions, e.g.
 ADD = 01001100.
 B. Problem. How does CPU distinguish
instructions and data?
 By position.
Instructions continued.
 INSTRUCTION:
 (Add what is in the variable located in 0100
to what is in the variable located in 0101)
 Operator
Operands
 01001100
0100 0101

[memory locations]
 See diagram.
Instructions continued.
 How does the compiler translate source
code into binary code?
 It uses a dictionary, which translates
HIGH-LEVEL instructions in a
programming language (e.g. ADD, PRINT)
into LOW-LEVEL binary instructions.