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Chapter 2
Bits, Data Types & Operations
Bits, Data Types & Operations
z Integer Representation
z Floating‐point Representation
z Other data types
Why do Computers use Base 2?
z
Base 10 Number Representation
– Natural representation for human transactions
z
H d I l
Hard to Implement Electronically
El
i ll
3.5 = 3×100 + 5×10-1
– Hard to store
z
ENIAC (First electronic computer) used 10 vacuum tubes / digit
– Hard to transmit
z
N d hi h
Need high precision to encode 10 signal levels on single wire
ii t
d 10 i l l l
i l i
– Messy to implement digital logic functions
z
z
Addition, multiplication, etc.
Binary Representation
– Specifically
Specifically, the devices that make up a computer are switches that can be the devices that make up a computer are switches that can be
on or off, i.e. at high or low voltage.
– Easy to store with bistable elements
– Reliably transmitted on noisy and inaccurate wires z
Examples
3.5 = 1×21 + 1×20 + 1×2-11 +…
– Represent 3.510 as 11.12
– Represent 1521310 as 111011011011012
– Represent 1.2010 as 1.0011001100110011[0011]…2
z Binary floating point number cannot exactly represent $1.20
Binary floating point number cannot exactly represent $1 20
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Data representation in binary
z
We need a way to represent information (data) in a form that is mutually comprehensible by human and machine.
that is mutually comprehensible by human and machine.
– We have to develop schemes for representing all conceivable types of information – integers, characters, floating point numbers, language, images, actions, etc.
z
Binary digITs (Bits): sequences of 0’s and 1’s that help humans keep track of the current flows stored/processed in the computer
– Using base‐2 provide us with two symbols to work with: we can call them on & off, or (more usefully) 0 and 1.
– We group bits together to allow representation of more complex information
l i f
ti
– For ease of use, Computer scientists usually represent quarters of bits in Hexadecimal
– eg. xA13F vs. 1010000100111111
eg xA13F vs 1010000100111111
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Hex
Binary
x0
0000
x1
0001
x2
0010
x3
0011
x4
0100
x5
0101
x6
0110
x7
7
0111
x8
1000
x9
1001
xA
1010
xB
1011
xC
1100
xD
1101
xE
1110
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xF
1111
3
Unsigned Binary Integers
Y = “abc” = a.22 + b.21 + c.20
(where the digits a, b, c can each take on the values of 0 or 1 only)
N = number of bits
Range is: 0 ≤ i < 2N – 1
Umin = 0
Umax = 2N – 1
Problem:
• How do we represent
g
numbers?
negative
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3-bits
5-bits
8-bits
0
000
00000
00000000
1
001
00001
00000001
2
010
00010
00000010
3
011
00011
00000011
4
100
00100
00000100
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Signed Magnitude
z
Leading bit is the sign bit
Y = “abc” = (-1)a (b.21 + c.20)
Range
g is:
-2N-1
+1<i <
2N-1
–1
Smin = -2N-1 + 1
Smax = 2NN-11 – 1
Problems:
• How do we do addition/subtraction?
• We have two numbers for zero (+/-)!
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-4
10100
-3
10011
-2
10010
-1
10001
-0
10000
+00
00000
+1
00001
+2
00010
+3
00011
+4
00100
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Two’s Complement
-16
10000
…
…
-3
11101
-22
11110
-1
11111
0
00000
Advantages:
+1
00001
• Operations need not check the sign
+2
00010
• Only one representation for zero
+3
00011
• Efficient use of all the bits
…
…
+15
01111
z
Transformation
– To transform a into ‐a, invert all bits in a and add 1 to the result
N 1 < i < 2N-1
N1–1
R
Range
i -22N-1
is:
Tmin = -2N-1
Tmax = 2N-1 – 1
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Manipulating Binary numbers ‐‐ 1
Manipulating Binary numbers z
Binary to Decimal conversion & vice‐versa
– A 4 bit binary number A = a3a2a1a0 corresponds to:
a3 x 23 + a2 x 22 + a1 x 21 + a0 x 20 = a3 x 8 + a2 x 4 + a1 x 2 + a0 x 1
(where ai = 0 or 1 only)
– A decimal number can be broken down by iteratively determining the highest power of two that “fits” in the number:
e.g. (13)10 =>
e.g. (63)10 =>
e.g. (0.63)10 =>
– In the 2’s complement representation, leading zeros do not affect the value of a positive binary number, and leading ones do not affect the value of a negative number. So:
01101 = 00001101 = 13 and 11011 = 11111011 = 5
01101 = 00001101 = 13 and 11011 = 11111011 = ‐5
00001101
11111011
00001000 => 8 (as expected!)
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x0D
xFB
x08
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Manipulating Binary numbers ‐‐ 10
Manipulating Binary numbers z
Overflow
– If we add the two (2’s complement) 4 bit numbers representing 7 and 5 we get :
0111
=> +7
0101
=> +5
5
1100
=> ‐4 (in 4 bit 2’s comp.)
z
z
z
We get ‐4, not +12 as we would expect !!
We have overflowed the range of 4 bit 2
We have overflowed
the range of 4 bit 2’ss comp. (‐8 to +7), so the result is invalid.
comp ( 8 to +7) so the result is invalid
Note that if we add 16 to this result we get back 16 ‐ 4 = 12
– this is like “stepping up” to 5 bit 2’s complement representation
– In general, if the sum of two positive numbers produces a negative result, or I
l if h
f
ii
b
d
i
l
vice versa, an overflow has occurred, and the result is invalid in that representation.
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CS Reality #1
z
z
The first place where the underlying hardware abstraction can’t be treated as a magic black box results in the need for data types
You’ve got to understand binary encodings.
– Can x = 20,000,000,000?
– Is x2 ≥ 0?
z
Int’s:
–
–
z
40000 * 40000 ‐‐> 1,600,000,000
50000 * 50000 ‐‐> ?? Fl t’ Yes!
Float’s: Y !
– Is (x + y) + z = x + (y + z)?
z
z
Unsigned & Signed Int’s: Float’s:
Float
s:
–
–
Yes!
(1e20 + ‐1e20) + 3.14 ‐‐> 3.14
1e20 + (‐1e20 + 3.14) ‐‐> 0? 3? 3.1?
– Is 4/5 = 4/5.0?
z
Maybe? (int) 4/5 = 0 what is 4/5.0?
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C++ Integer Puzzles
– Assume machine with 32 bit word size, two’s complement integers
– For each of the following C++ expressions, either:
For each of the following C++ expressions either
z
z
Argue that is true for all argument values
Give example where not true
• x * x >= 0
• ux >= 0
Initialization
• ux > -1
1
int x = foo();
• x < 0
⇒ (2*x) < 0
int y = bar();
• x > y
⇒ -x < -y
unsigned ux = x;
• x > 0 && y > 0
⇒ x + y > 0
unsigned uy = y;
• x >= 0
⇒ -x <= 0
• x <= 0
⇒ -x >= 0
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Problems
z
Patt & Patel, Chapter 2:
p
– 2.4, 2.8, 2.10, 2.11, 2.17, 2.21
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Representing Strings
z
Strings in C/C++
– Represented by array of characters
– Each character encoded in ASCII format
z
z
Standard 7‐bit encoding of character set
Other encodings exist, but uncommon
– UNICODE
UNICODE 16
16‐bit
bit superset of ASCII that superset of ASCII that
includes characters for non‐English alphabets
z
Character “0” has code 0x30
– Digit i has code 0x30+i
char S[6] = "15213";
x00A6
x00A7
x00A8
x00A9
x00AA
x00AB
– String should be null‐terminated
z
x31
x35
x32
x31
x33
x00
#49
#53
#50
#49
#51
#00
Hex
Dec
Final character = 0
– Line termination, and other special characters vary
characters vary
Strings are really just arrays of numbers!
How can we represent numbers using bits?
“Hey
Hey, how do I know it
how do I know it’s a string in the first place?”
s a string in the first place?
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Real numbers
z
z
Most numbers are not integer!
Range:
– The magnitude of the numbers we can represent is determined by how many bits we use:
z
z
e.g. with 32 bits the largest number we can represent is about +/‐ 2 billion, far too small for many purposes.
Precision:
– The exactness with which we can specify a number:
z
z
e.g. a 32 bit number gives us 31 bits of precision, or roughly 9 figure precision in decimal representation representation
Our decimal system handles non‐integer real numbers by adding yet another symbol ‐ the decimal point (.) to make a fixed point notation:
3 + 5.102 + 4.101 + 6.100 + 7.10‐1 + 8.10‐2
– e.g. 3,456.78 = 3.10
g ,
z
The floating point, or scientific, notation allows us to represent very large and very small numbers (integer or real), with as much or as little precision as needed.
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Real numbers in a fixed space
z
z
What if you only have 8 digits to represent a real number?
Do you prefer…
... a low‐precision number with a large range?
… or a high‐precision number with a smaller range?
Wouldn’t it be nice if we could “float” the decimal f
f
point to allow for either case?
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Scientific notation
z
z
z
In binary, we only have 0 and 1, how can we represent the decimal point?
i ?
In scientific notation, it’s implicit. In other words:
425000 (× 100) =
425 × 103 =
42.5 × 104 =
4.25 × 105
4.25 ×
We can represent any number with the decimal after the first digit by “floating” the decimal point to the left (or right)
0 00125 =
0.00125 =
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Real numbers in binary z
We mimic the decimal floating point notation to create a “hybrid” binary floating point number:
floating point number:
– We first use a “binary point” to separate whole numbers from fractional numbers to make a fixed point notation:
z
e.g. 25.75
e.g.
25.7510 1.24 + 1.10
1.103 + 1.10
1.101 + 1.2
1.2‐1 + 1.2
1.2‐2 = 11001.1102
10 = 1.2
(2‐1 = 0.5 and 2‐2 = 0.25, etc.)
– We then “float” the binary point:
z
00011001.110 => 1.1001110 x 24
mantissa = 1.1001110, exponent = 4
– Now we have to express this without the extra symbols (
p
y
( x, 2, . )
, , )
z
by convention, we divide the available bits into three fields:
sign, mantissa, exponent
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IEEE‐‐754 IEEE
754 fp
fp numbers numbers –– 32 bit “float”
32 bits:
1
8 bits
s biased exp.
23 bits
fraction
N = (-1)s x 1.fraction x 2(biased exp. – 127)
z
Sign: 1 bit
Sign: 1 bit
z
Mantissa: 23 bits
– We “normalize” the mantissa by dropping the leading 1 and recording only its f ti
fractional part (why?) l
t ( h ?)
z
Exponent: 8 bits
– In order to handle both + and ‐ exponents, we add 127 to the actual exponent to create a “biased exponent” (this provides a lexographic order!) tt
t “bi d
t” (thi
id
l
hi
d !)
z
z
z
2‐127 => biased exponent = 0000 0000 (= 0)
20 => biased exponent = 0111 1111 (= 127)
2+127 =>> biased exponent biased exponent = 1111 1110 (
1111 1110 (= 254)
254)
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IEEE‐‐754 IEEE
754 fp
fp numbers numbers ‐‐ example
z
z
Example:
z
25.75 =>> 00011001.110 25.75 00011001.110 =>> 1.1001110 x 2
1.1001110 x 24
z
sign bit = 0 (+)
z
normalized mantissa (fraction) = 100 1110 0000 0000 0000 0000
z
biased exponent = 4 + 127 = 131 => 1000 0011
biased exponent = 4 + 127 = 131 => 1000 0011
z
so 25.75 => 0 1000 0011 100 1110 0000 0000 0000 0000 => x41CE0000
Special values represented by convention:
– Infinity (+ and ‐): exponent = 255 (1111 1111) and mantissa = 0
– NaN (not a number): exponent = 255 and mantissa ≠
NaN (not a number): exponent = 255 and mantissa ≠ 0
– Zero (0): exponent = 0 and mantissa = 0
z
note: special case => fraction is de‐normalized, i.e no hidden 1
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IEEE‐‐754 IEEE
754 fp
fp numbers numbers –– 64
64‐‐bit “double”
Double precision (64 bit) floating point
z
64 bits:
1
11 bits
s biased exp.
52 bits
fraction
N = (-1)s x 1.fraction x 2(biased exp. – 1023)
z
R
Range
&P
Precision:
i i
Š 32 bit:
ƒ mantissa of 23 bits + 1 => approx. 7 digits decimal
+/-127
127 => approx.
+/-38
38
ƒ 2+/
approx 10+/
Š 64 bit:
ƒ mantissa of 52 bits + 1 => approx. 15 digits decimal
+/ 1023 => approx. 10+/-306
+/ 306
ƒ 2+/-1023
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Floating point in C/C++
z
C Guarantees Two Levels
float
double
z
single precision
double precision
Conversions
– Casting between int, float, and double
g
,
,
changes numeric values
g
– Double or float to int
z
z
z
Truncates fractional part
Like rounding toward zero
Not defined when out of range
– Generally saturates to TMin or TMax
– int to double
z
E
Exact conversion, as long as int has ≤
i
l
i h ≤ 53 bit word size
53 bi
d i
– int to float
z
Will round according to rounding mode
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Floating point puzzles in C++
int x = …;
float f = …;
;
Assume neither
d nor f is NAN
double d = …;
• x == (int)(float) x
• x == (int)(double) x
• f == (float)(double) f
• d == (float) d
• f == -(-f);
• 2/3 == 2/3.0
0.0
0 ⇒ ((d
((d*2)
2) < 0.0)
0 0)
• d < 0
• d > f
⇒ -f > -d
• d * d >= 0.0
• (d+f)-d
(d f) d == f
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Another floating point puzzle
#include <iostream>
using namespace std;
int main() {
float i;
for (i=0.0;i<25.0;i+=0.1) {
cout << i << endl;
}
}
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[...]
7
7.1
7.2
7.3
7.4
7.5
7.6
7 7
7.7
7.79999
7.89999
7.99999
8.09999
8.2
8.3
[
[...]
]
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Floating point number line
z
z
z
32 bits can represent 232 unique values
How are those values distributed differently between integers and y
g
floats?
What are the implications?
-1
0
+1
-10
0
+1
0
-2b
0
+2b
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Ariane 5
z
z
z
Payload rocket
7 billion (Euro) in design alone
(
)
g
Danger!
– Computed horizontal velocity gp
as floating point number
– Converted to 16‐bit integer
– Worked OK for Ariane 4
– Overflowed for Ariane 5
z
z
same software
Result
– Exploded 37 seconds after xploded 37 seconds after
liftoff
– Cargo worth $500 million
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Other Data Types
z
Other numeric data types
– e.g. BCD
BCD
z
Bit vectors & masks
– sometimes we want to deal with the individual bits themselves
z
Logic values
– True (non‐zero) or False (0)
z
Instructions
– Output as “data” from a compiler
z
Misc
– Grapics, sounds, …
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Practice Problems
z
Patt & Patel
– 2.39, 2.40, 2.41, 2.42, 2.56
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