Download 1 - Electronic Systems Group

Computation
Course 5JJ70 “Implementatie van
rekenprocessen”
Henk Corporaal
p
September 2009
h corporaal@tue nl
[email protected]
Welcome to 5JJ70 !
• 5JJ70 is the biggest course in the 1st semester. It deals with
the hardware and software of a computing system:
– Electronics
• on the elementary components of a computer.
– Computer
p
architecture
• on the structure of computer hardware.
– The hardware-software interface
• on software that runs on the hardware.
My contribution
• Computer / Embedded System:
System
– “Microprocessor”, “Ordinateur”, “EDV-Anlage"
((Elektronische Datenverarbeitungsanlage),
g
g )
“Elaboratore elettronico”
• This is the hart of any:
– PC, PDA, fly-by-wire
y y
system,
y
MP3-player,
p y
Mobile
phone,
h
Smart
S
t card,
d Smart
S
t bomb,
b b TV
TV, W
Washing
hi
machine, car engine, modem, etc.
5JJ70 Programming pg 2
Programming in C and C++
• We will learn how computer software really ‘runs’ on the
hardware:
– Machine language,
g g , instructions,, data paths
p
• How to program in C and C++:
– The languages
g g of
f real p
programmers.
g mm
• At the end, you can attack virtually any real-life programming
task!
• This is intensive: you’ll learn a lot in a very short time:
– Basic operations
– Pointers,
Pointers dynamic memory allocation
– Machine language
– Data structures, Object-oriented programming
– Cool algorithms
5JJ70 Programming pg 3
The schedule : 1th semester
• Classes (‘college’):
(‘college’) 4 hours per week
– Tuesday,
– Friday,
Friday
– In EH 1.01
hours 3+4 (10.45)
hours 3+4 (10
(10.45)
45)
• Labs (“practicum”): 4 hours per week
– Monday:
hours 1+2
– Thursday:
hours 7+8
– In EH6.05 and EH6.23
– You work on your laptop
• Lab Assisted by
– Raymond Frijns
j
– Sven Goossens
– Emil Kleijn
5JJ70 Programming pg 4
The labs (“Practicum”)
• Programming tasks:
– Handed out in the lab. Take y
your laptop.
p p
– You work individually on your programming assignments.
– Discussing or asking your neighbors is OK, copying is not OK.
– You will have to sign
g off each assignment
g
with the staff.
• We check whether you did it yourself.
– Hand-in your assignment, using peach: peach.win.tue.nl
• We’ll check for plagiarism
– While working on an assignment you can take breaks as you please.
– Assignments must be finished (signed off and emailed) within 7
days after handout.
handout
• Lab assistents will sign off and grade your assignments.
5JJ70 Programming pg 5
Computer revolution
P
Performa
ance
• Computers everywhere
– PCs, portables, and workstations
O(100M)/year
– Embedded systems
O(10G)/year
– Playstations
O(100M/year)
• World Wide Web
• Huge calculations
– Human genome project
– Huge simulations (e.g. weather forecast)
• Information retrieval
– Huge databases
Processing
– Online libraries
– Web pages
55 %/y
%/
Performance
• Enormous computer
gap
performance growth
Memory
35 %/y
7 %/y
70
80
90
00
10
Year
5JJ70 Programming pg 6
What is a computer system?
Display
harddisk
network
connection
printer
keyboard
Q What is an
Q.
embedded system ?
actuators
sensors
5JJ70 Programming pg 7
Embedded system: ADSL modem
1. Telephone decoupling
2/3. LEDs
4. Main (TI) processor
5. JTAG test port
6 RAM,
6.
RAM 8 MB chip.
hi
7. Flash memory
8. Power regulator.
9 fuse.
9.
fuse
10. Power connector
11. Reset button
12 Quartz crystal
12.
13. Ethernet port
14. Ethernet transformer
15. Ethernet PHY transmitter receiver
16. USB
17. Telephone (RJ11)
18. Telephone connector fuses.
5JJ70 Programming pg 8
Conceptual structure of a computer system;
How does it operate?
Instruction
memory
Data
memory
Processor
I/O
i t f
interfaces
Input and output
devices
5JJ70 Programming pg 9
Binary representations
• 0110 0101 0100 0100 1110 0111 0110 1100 0110 0101 0100
0100 1110 0111 0110 1100 0110 0101 0100 0100 1110 0111
0110 1100 0110 1100 0110 0101 0100 0100 1110 0111 0110
•1100
Computers
p
contain
many
y megabytes
g y1110 of
memory.
y
0110 1100
0110 0101
0100
0100
0111
0110 1100
0110 1100 0110 0101 0100 0100 1110 0111 0110 1100 0110
1100 0110 0101 0100 0100 1110 0111 0110 1100 0110 1100
•0110
Without
conventions
completely
0101 0100
0100 1110this
0111data
0110 is
1100
0110 1100 0110
0101
0100 0100 1110 0111 0110 1100 0110 1100 0110 0101
meaningless
meaningless.
0100 0100 1110 0111 0110 1100 0110 1100 0110 0101 0100
0100 1110 0111 0110 1100 0110 1100 0110 0101 0100 0100
0111 0110
1100
1100
0101 data,
0100 0100 1110
•1110
We must
agree
on0110
a way
to 0110
encode
data
0111
0110 1100numbers.
0110 1100 0110 0101 0100 0100 1110 0111
particularly
0110 1100 0110 1100 0110 0101 0100 0100 1110 0111 0110
1100 0110 1100 0110 0101 0100 0100 1110 0111 0110 1100
1100 0110 0101 0100 0100 1110 0111 0110 1100 0110
•0110
Th
The
good
d news: we have
h
lots
l
of
f bytes
b
in
i today’s
d ’
1100 0110 0101 0100 0100 1110 0111 0110 1100 0110 1100
computers.
0110
0101 0100 0100 1110 0111 0110 1100 0110 1100 0110
0101 0100 0100 1110 0111 0110 1100 0110 1100 0110 0101
0100 0100 1110 0111 0110 1100 0110 1100 0110 0101 0100
0100 1110 0111 0110 1100 0110 1100 0110 0101 0100 0100
1110 0111 0110 1100
5JJ70 Programming pg 10
Computer structure
CPU
(e.g. pentium)
Address
(e.g. 32 bit)
Data (e.g.
(e g 32 bit)
Memory
(e.g. 5
512
Megabyte Sims =
128 M data words)
This is a memory read
D i write
During
it the
th d
data
t goes ffrom th
the CPU into
i t the
th memory
• 32 bit data = 0110 0101 0100 0100 1110 0111 0110 1100
• What could this data represent?
– A number (E.g. Your Postgiro account balance)
– An instruction for the CPU (e.g. add two numbers)
– 4 ASCII text characters of 8 bits each
– 0.25 milliseconds of MP3 music or 0.045 milliseconds of CD music
– Some (image) pixels, e.g. a piece of your book cover
5JJ70 Programming pg 11
Representing decimal numbers
• Human calculation is mainly in the decimal number system.
We use 10 symbols:
0123456789
• We call this ‘base 10’
– Why did mankind pick base 10?? (and not base 8?)
• We can encode larger numbers by strings of numbers:
1969
1000
900
60
9
=
+ (= 1 *
+ (= 9 *
+ (= 6 *
(= 9 *
(
10^3)
10^2)
10^1)
10^0)
)
5JJ70 Programming pg 12
Bases
• We can use an arbitrary base g. Then the value of of digit c at
position i is:
valuei = c ∗ g
i
– (Note that we start counting digit positions at 0 )
• For example,
example in the decimal system g = 10.
10 Digit 3 in 1969 is 1,
1 and
therefore its value is:
value3 = 1 ∗103 = 1000
• In this way we can represent numbers using an arbitrary base.
Popular base values (g) are:
–g =
–g =
–g =
–g =
2: BINARY numbers
8: OCTAL numbers
10: DECIMAL number
16: HEXADECIMAL number
e.g. 1001
e.g. 3661
e
e.g.
g 1969
e.g. 4BF7
(= decimal 9)
5JJ70 Programming pg 13
Binary numbers
• Computer hardware is binary:
– the base is 2, with only two symbols: 0 and 1
(why?)
• Converting from binary to decimal:
n −1
(an −1 an −2 ... a2 a1 a0 )two = ∑ ai ∗ 2i =
i =0
an −1 ∗ 2 n −1 + an − 2 ∗ 2 n − 2 + ... + a2 ∗ 2 2 + a1 ∗ 21 + a0 ∗ 2 0 =
Least significant bit: LSB
Most significant bit: MSB
n − 1 n − 2 ...
1
0
Bit position
an −1
an − 2 ...
a1
a0
digit
g
2 n −1
2 n − 2 ...
21
20
weight
5JJ70 Programming pg 14
With n bits 2n possible combinations
Binary (g = 1)
1 bit
0
1
2 bits
00
01
10
11
3 bits
000
001
010
011
100
101
110
111
4 bits
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Decimal
(g = 10))
value
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Octal
(g = 8)
value
0
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
Hexadecimal
(g = 16)
value
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
5JJ70 Programming pg 15
Bits, nibbles and bytes
1
1010
10000001
1111111111001110
11111111110011101111111111001110
• 1 bit
• 4 bits = nibble (‘knabbel’)
• 8 bits = byte
• 16 bits (C: short int)
• 32 bits ‘word’
Using n bits, we can represent positive integer numbers
between 0 and 2n -1
[0,2 n
5JJ70 Programming pg 16
Representing characters: 1 byte encoding
A
American
i
St d d Code
Standard
C d ffor Information
I f
ti Interchange
I t h
5JJ70 Programming pg 17
Conversion from decimal to binary
• Divide by 2
4382
2191
1095
547
273
136
68
34
17
8
4
2
1
0
Remainder
0
1
1
1
1
0
0
0
1
0
0
0
1
4382ten =
1 0001 0001 1110bin
What's the trick ?
• Conversion between hexadecimal (base 16) and binary is simple:
– 1010101100111111bin = ab3fhex
5JJ70 Programming pg 18
Pop Quiz!
• We need to store the (decimal) number 2500000000 (2.5
Billion) as a binary number.
– What is the minimum number of bits that are required to
store this (unsigned) number?
• Now we store the same number as a string of ASCII
characters.
– How many bits are required?
5JJ70 Programming pg 19
Operating on binary numbers
• Addition is similar to decimal
– starting from the LSB, add two bits
– if the sum is larger
g or equal
q
to base g (=2),
( ) produce
p
a carry:
y
1 1111
00011111
10111010
1 10110 01
bin
bin
+
31
186 +
217dec
• Multiplication/division is also familiar
– Multiplication by a power of two is a ‘left shift’:
00011111bin * 10bin = 00111110bin
00111000bin / 10bin = 00011100bin
5JJ70 Programming pg 20
Multiplication: example
• Multiplication is left shift followed by an addition:
10 X 5
1010
101
bin
1010
00000
101000
bin
X
1 * 1010
00 * 1010 (=Shift left 1)
+
100 * 1010 ((=Shift
f left
f 2))
110010
5JJ70 Programming pg 21
More on binary number
representations
• binary numbers (base 2)
0000 0001 0010 0011 0100 0101 0110 0111 1000 1001...
1001
N digits = 2n numbers
representing decimal: 0...2n-1
• We have to deal with a few problems:
1) numbers are finite (overflow due to finite range)
2) fractions and real numbers
3) negative numbers
4) complex numbers, etc….
how do we represent negative numbers?
i.e., which bit patterns will represent which numbers?
5JJ70 Programming pg 22
Signed binary numbers
A few possible representations:
sign magnitude:
000
001
010
011
100
101
110
111
=
=
=
=
=
=
=
=
+0
+1
+2
+3
-0
-1
1
-2
-3
one
one'ss complement:
000
001
010
011
100
101
110
111
=
=
=
=
=
=
=
=
+0
+1
+2
+3
-3
-2
2
-1
-0
x
two's
two s complement:
000 = +0
001 = +1
010 = +2
011 = +3
100 = -4
101 = -3
3
110 = -2
111 = -1
xˆ = x + 1
• Issue: ease of operations, compactness
5JJ70 Programming pg 23
Getallencirkel: Sign magnitude
subtraction
1000
1010
1011
0011
0101 ????
0000
1001
0001
0
-1
1
3 - 5 = -2
0010
-2
2
-3
-(5
(5 - 3) = -2
2
3
0011
Swap 5 and 3
1100
1101
iti
positive
-4
4
negative
-55
5
-6
6
1110
-7
1111
0110
7
0111
0100
0101
101
011 010
1
1010
Add sign
(‘getallencirkel’)
5JJ70 Programming pg 24
The ‘getallencircle’, two’s complement
0000
1111
1110
0001
-1
15
-2
0
1
0010
2
14
1101
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
+0
+1
+2
+3
+4
+5
+6
+7
-8
-7
-6
-5
-4
-3
-2
-1
1100
1011
-3
3
13
positive
-4 12
4
negative
-5
0011
0100
11
5
0101
10
-6
1010
1001
9
-7
6
8
-8
0110
7
0111
1000
5JJ70 Programming pg 25
4-bit two’s complement
(a3 a2 a1 a0) 2’s-compl =
1111
-a3 23 + a2 22 + a1 21 + a0 20
1110
0001
-1
15
-2
0
1
0010
2
14
1101
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
+0
+1
+2
+3
+4
+5
+6
+7
-8
-7
-6
-5
-4
-3
-2
-1
1100
1011
3+2=5
0000
-3
3
13
positive
-4 12
-5
0011
4
negative
0011
0010 +
0101bin
0100
11
5
0101
10
-6
1010
1001
9
-7
6
8
-8
0110
7
0111
1000
Range = [-2 3 .. 2 3 -1]
5JJ70 Programming pg 26
Two’s complement subtraction
0000
1111
1110
0
1
0010
2
14
1101
0011
1011 +
1110bin
-1
15
-2
3 -5 = 3 + (-5) = -2
0001
1100
1011
-3
3
13
positive
-44 12
4
negative
-5
0011
0100
11
5
0101
10
-6
1010
1001
9
-77
6
8
-8
0110
7
0111
1000
5JJ70 Programming pg 27
Two’s complement add: overflow possible!
0000
0011
0110 +
1001bin
1111
1110
0001
-1
15
-2
1
0010
2
14
1101
1100
1011
-3
3+6 = 9
0
3
13
positive
-4 12
4
negative
-5
0011
0100
11
5
0101
10
-6
1010
1001 = -7
7
overflow!!!
1001
9
-7
6
8
-8
1000
0110
7
0111
Overflow
O
fl
if
we cross here
5JJ70 Programming pg 28
Two’s complement subtract: overflow also
possible!
0000
1101
1010 +
0111bin
1111
1110
-1
15
-2
0
1
0010
2
14
1101
1100
1011
-3
-3 + (-6) = 7 !!
underflow
0001
3
13
positive
-4 12
4
negative
-5
0011
0100
11
5
0101
10
-6
1010
1001
9
-7
6
8
-8
0110
7
0111
1000
5JJ70 Programming pg 29
Two’s complement: C integer
• 32 bit signed numbers:
0000
0000
0000
...
0111
0111
1000
1000
1000
...
1111
1111
1111
maxint
0000 0000 0000 0000 0000 0000 0000two = 0ten
0000 0000 0000 0000 0000 0000 000
0001two = + 1ten
0000 0000 0000 0000 0000 0000 0010two = + 2ten
1111
1111
0000
0000
0000
1111
1111
0000
0000
0000
1111
1111
0000
0000
0000
1111
1111
0000
0000
0000
1111
1111
0000
0000
0000
1111
1111
0000
0000
0000
1110two
1111two
0000two
0001two
0010two
=
=
=
=
=
+
+
–
–
–
minint
2,147,483,646ten
2,147,483,647ten
2,147,483,648
,
,
,
ten
2,147,483,647ten
2,147,483,646ten
1111 1111 1111 1111 1111 1111 1101two = – 3ten
1111 1111 1111 1111 1111 1111 1110two = – 2ten
1111 1111 1111 1111 1111 1111 1111two = – 1ten
– Range [-2
31
.. 2
31
-1]
• (an-1 an-2... a1 a0) 2’s-compl = -an-1 x 2n-1 + an-2 x 2n-2 + … + a0 x 20
5JJ70 Programming pg 30
Addition & subtraction: summary
• just like in grade school (carry/borrow 1s)
0111
0111
0110
+ 0110
- 0110
- 0101
• two's complement operations easy
– subtraction
b
i using
i addition
ddi i of
f negative
i numbers
b
0110
0110
- 0101
+ 1011
• overflow (result too large for finite computer word):
– e.g., adding two n-bit numbers does not yield an n-bit
number
0111
+ 0001 ,
1000
5JJ70 Programming pg 31
Two's complement operations
• Negating a two's complement number: invert all bits and
add 1..
a
– remember: “negate” and “invert” are quite different
• Example (50 negated into -50):
50):
00110010
11001101
00000001 +
invert
Add 1
11001110bin
5JJ70 Programming pg 32
Two's complement: word length change
Processors often load 8 or 16 bit numbers into 32 bit
registers for operations. How do we keep convert the
word length?
• The trick: copy the most significant bit (the sign bit) into
the empty bits. For example 8 bit converted into 16 bit:
•
11001110
00110010
1111111111001110
0000000000110010
• Some
processors have
special
for
S
h
i l machine
hi instructions
i t
ti
f
this. E.g. MIPS "sign extension” instructions:
•
•
•
•
Lb
Lbu
Slti
Sltiu
load byte (signed)
load byte (unsigned)
set less than immediate (signed)
set less than immediate (unsigned)
5JJ70 Programming pg 33
Detecting overflow
• Never overflow when adding a positive and a negative
number.
• No overflow when signs are the same for subtraction
• Overflow occurs when the value affects the sign:
– overflow when adding two positives yields a negative
– or,
or adding two negatives gives a positive
– or, subtract a negative from a positive and get a negative
– or, subtract a positive from a negative and get a positive
5JJ70 Programming pg 34
Pop quiz!
• We have two numbers in two’s complement: a and b.
Consider the operations
– Can overflow occur if b == 0
– Can overflow occur if a == 0
000
001
010
011
100
101
110
111
=
=
=
=
=
=
=
=
a + b and a - b
?
?
?
?
+0
+1
+2
+3
-4
-3
3
-2
-1
5JJ70 Programming pg 35
Summary
• Computers encode numbers binary, typically in 4, 8, 16,
32 64 or 128 bit
32,
• Signed number are encoded using ‘two’s complement’
• We can only encode a finite set of numbers. In any case,
we mustt be
b careful
f l for
f overflow,
fl
i case th
in
the result
lt
cannot be encoded in the given amount of bits.
Material: Patterson Hennessy: Section 2.4
Ward/Halstead: 2.1 - 2.5
5JJ70 Programming pg 36