Transcript 1 - La Salle University

Recap so far (Not gotten to last time)
• So there were issues about the number of
operands.
– Recall that we have a fetch-execute cycle – first an
instruction is retrieved from memory and then acted
upon.
– With unary instructions adding two numbers and
storing the result required three instructions, that’s three
fetches and three executions.
– With ternary instructions it can be done with one
instruction, one fetch and one execute. The execution is
now more complicated but we have saved time on
fetches.
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Recap so far (Cont.)
• More operators means more complicated
circuitry, the load and store aspects of the
instruction would have to built into each
separate instruction.
• There is a speed versus complexity issue.
And complexity also brings the issue of cost
along with it.
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Recap so far (Cont.)
• After determining the number of operands, came
the issue of what the operands mean.
– Are they data, addresses of data, or addresses of
addresses of data?
• Either we can decide to support all of these types
of instructions (addressing modes) and choose
complexity. Or we can choose to support only
some of them and sacrifice efficiency.
– You can eliminate Add Immediate if you always store
the values you want to add.
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Data Types
• Apart from addressing, another issue is the type of
data the operation is acting on.
– The process for adding integers is different from the
process for adding floating point numbers.
– So one may have separate ADD and FADD for the
addition of integers and floats respectively.
– Furthermore, one may need to add instructions to
convert from one type to another.
• To add an integer to a float, convert the integer to a float and
then add the floats.
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Binary Numbers
Material on Data Representation can be found in
Chapter 2 of Computer Architecture (Nicholas Carter)
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Why Binary?
• Maximal distinction among values 
minimal corruption from noise.
• Imagine taking the same physical attribute
of a circuit, e.g. a voltage lying between 0
and 5 volts, to represent a number.
• The overall range can be divided into any
number of regions.
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Don’t sweat the small stuff
• For decimal numbers, fluctuations must be less
than 0.25 volts.
• For binary numbers, fluctuations must be less
than 1.25 volts.
5 volts
0 volts
Decimal
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Binary
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Range actually split in three
High
Forbidden
range
Low
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It doesn’t matter ….
• Two of the standard voltages coming from a
computer’s power supply are ideally
supposed to be 5.00 volts and 12.00 volts
• Measurements often reveal values that are
slightly off – e.g. 5.14 volts or 12.22 volts
or some such value.
• So what, who cares.
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How to represent big integers
• Use positional weighting, same as with
decimal numbers
• 205 = 2102 + 0101 + 5100
• 11001101 = 127 + 126 + 025 + 024
+ 123 + 122 + 021 + 120
= 128 + 64 + 8 + 4 + 1
= 205
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Converting 205 to Binary
• 205/2 = 102 with a remainder of 1, place the
1 in the least significant digit position
1
• Repeat 102/2 = 51, remainder 0
0
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Iterate
• 51/2 = 25, remainder 1
1
0
1
1
0
1
1
0
1
• 25/2 = 12, remainder 1
1
• 12/2 = 6, remainder 0
0
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Iterate
• 6/2 = 3, remainder 0
0
0
1
1
0
1
1
1
0
1
1
1
0
1
• 3/2 = 1, remainder 1
1
0
0
• 1/2 = 0, remainder 1
1
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1
0
0
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Recap
1
1
0
0
1
1
0
1
127 + 126 + 025 + 024
+ 123 + 122 + 021 + 120
205
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Finite representation
• Typically we just think computers do binary math.
• But an important distinction between binary math
in the abstract and what computers do is that
computers are finite.
• There are only so many flip-flops or logic gates in
the computer.
• When we declare a variable, we set aside a certain
number of flip-flops (bits of memory) to hold the
value of the variable. And this limits the values
the variable can have.
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Same number, different
representation
•
•
•
•
•
•
5 using 8 bits
0000 0101
5 using 16 bits
0000 0000 0000 0101
5 using 32 bits
0000 0000 0000 0000 0000 0000 0000 0101
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Adding Binary Numbers
• Same as decimal; if the sum of digits in a
given position exceeds the base (10 for
decimal, 2 for binary) then there is a carry
into the next higher position
+
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3
3
7
9
5
4
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Adding Binary Numbers
carries
1
+
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1
1
1
0
1
0
0
1
1
1
39
0
1
0
0
0
1
1
35
1
0
0
1
0
1
0
74
18
Uh oh, overflow
• What if you use a byte (8 bits) to represent an
integer
1
1
1
1
0
1
0
1
0
1
0
170
1
1
0
0
1
1
0
0
204
0
1
1
1
0
1
1
0
118
???
• A byte may not be enough to represent the sum of
two such numbers.
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Biggest
*
unsigned
integers
4 bit: 1111  15 = 24 - 1
8 bit: 11111111  255 = 28 – 1
16 bit: 1111111111111111  65535= 216 – 1
32 bit: 11111111111111111111111111111111 
4294967295= 232 – 1
• Etc.
*If one uses all of the bits available to represent only
positive counting numbers, one is said to be working
with unsigned integers.
•
•
•
•
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Bigger Numbers
• High-level languages often offer a hierarchy of
types that differ in the number of bits used.
• You can represent larger numbers than allowed by
the highest type in the hierarchy by using more
words.
• You just have to keep track of the overflows to
know how the lower numbers (less significant
words) are affecting the larger numbers (more
significant words).
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Negative numbers
• Negative x is the number that when added to x
gives zero
1
1
1
1
1
1
1
1
0
0
1
0
1
0
1
0
 x
1
1
0
1
0
1
1
0
 -x
0
0
0
0
0
0
0
0
• Ignoring overflow the two eight-bit numbers
above add up to zero
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Two’s Complement: a two-step
procedure for finding -x from x
0
0
1
0
1
0
1
0
0
1
 x
• Step 1: exchange 1’s and 0’s
1
1
0
1
0
1
• Step 2: add 1 (to the lowest bit only)
1
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1
0
1
0
1
1
0
 -x
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Sign bit
• With the two’s complement approach, all
positive numbers start with a 0 in the leftmost, most-significant bit and all negative
numbers start with 1.
• So the first bit is called the sign bit.
• But note you have to work harder than just
strip away the first bit.
• 10000001 IS NOT the 8-bit version of –1
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Add 1’s to the left to get the same
negative number using more bits
•
•
•
•
•
•
•
-5 using 8 bits
11111011
-5 using 16 bits
1111111111111011
-5 using 32 bits
11111111111111111111111111111011
When the numbers represented are whole numbers
(positive or negative), they are called integers.
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3-bit signed and unsigned
7
6
5
4
3
2
1
0
1
1
1
1
0
0
0
0
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1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
3
2
1
0
-1
-2
-3
-4
0
0
0
0
1
1
1
1
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
Think of
driving a
brand new
car in
reverse.
What would
happen to the
odometer?
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Biggest signed integers
4 bit: 0111  7 = 23 - 1
8 bit: 01111111  127 = 27 – 1
16 bit: 0111111111111111  32767= 215 – 1
32 bit: 01111111111111111111111111111111
 2147483647= 231 – 1
• Etc.
•
•
•
•
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Most negative signed integers
4 bit: 1000  -8 = - 23
8 bit: 10000000  - 128 = - 27
16 bit: 1000000000000000  -32768= - 215
32 bit:
10000000000000000000000000000000 
-2147483648= - 231
• Etc.
•
•
•
•
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Riddle
1
•
•
•
•
•
1
0
1
0
1
1
0
Is it 214?
Or is it – 42?
Or is it Ö?
Or is it …?
It’s a matter of interpretation
– How was it declared?
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Hexadecimal Numbers
• Even moderately sized decimal numbers end
up as long strings in binary.
• Hexadecimal numbers (base 16) are often
used because the strings are shorter and the
conversion to binary is easier.
• There are 16 digits: 0-9 and A-F.
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Decimal  Binary  Hex
•
•
•
•
•
•
•
•
0  0000  0
1  0001  1
2  0010  2
3  0011  3
4  0100  4
5  0101  5
6  0110  6
7  0111  7
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•
•
•
•
•
•
•
8  1000  8
9  1001  9
10  1010  A
11  1011  B
12  1100  C
13  1101  D
14  1110  E
15  1111  F
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Binary to Hex
• Break a binary string into groups of four
bits (nibbles).
• Convert each nibble separately.
1 1 1 0 1 1 0 0 1 0 0 1
E
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Numbers from Logic
• All of the numerical operations we have talked
about are really just combinations of logical
operations.
• E.g. the adding operation is just a particular
combination of logic operations
• Possibilities for adding two bits
–
–
–
–
0+0=0 (with no carry)
0+1=1 (with no carry)
1+0=1 (with no carry)
1+1=0 (with a carry)
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Addition Truth Table
INPUT
OUTPUT
A
B
Sum
A XOR B
0
0
0
0
0
1
1
0
1
0
1
0
1
1
0
1
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Carry
A AND B
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Multiplication: Shift and add

+
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1
1
0
0
1
1
0
0
0
1
1
1
0
0
1
0
0
0
0
0
0
0
1
0
1
1
1
1
shift
shift
0
1
1
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Fractions
• Similar to what we’re used to with decimal
numbers
3.14159 =
3 · 100 + 1 · 10-1 + 4 · 10-2 + 1 · 10-3
+ 5 · 10-4 + 9 · 10-5
1 · 21 + 1 · 20 + 0 · 2-1 + 0 · 2-2
+ 1 · 2-3 + 0 · 2-4 + 0 · 2-5
+ 1 · 2-6
(11.001001  3.140625)
11.001001 =
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Converting decimal to binary II
• 98.61
– Integer part
•
•
•
•
•
•
•
98 / 2
49 / 2
24 / 2
12 / 2
6/2
3/2
1/2
= 49
= 24
= 12
= 6
= 3
= 1
= 0
remainder
remainder
remainder
remainder
remainder
remainder
remainder
0
1
0
0
0
1
1
– 1100010
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Converting decimal to binary III
• 98.61
– Fractional part
•
•
•
•
•
•
0.61  2 = 1.22
0.22  2 = 0.44
0.44  2 = 0.88
0.88  2 = 1.76
0.76  2 = 1.52
0.52  2 = 1.04
– .100111
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Another Example (Whole number part)
• 123.456
– Integer part
•
•
•
•
•
•
•
123 / 2 = 61 remainder 1
61 / 2 = 30 remainder 1
30 / 2 = 15 remainder 0
15 / 2 = 7 remainder 1
7 / 2 = 3 remainder 1
3 / 2 = 1 remainder 1
1 / 2 = 0 remainder 1
– 1111011
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Checking: Go to
Programs/Accessories/Calculator
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Put the calculator in Scientific view
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Enter number while in decimal mode,
then put Calculator into binary mode
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Another Example (fractional part)
• 123.456
– Fractional part
•
•
•
•
•
•
•
•
0.456  2 = 0.912
0.912  2 = 1.824
0.824  2 = 1.648
0.648  2 = 1.296
0.296  2 = 0.592
0.592  2 = 1.184
0.184  2 = 0.368
…
– .0111010…
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Checking fractional part: Enter digits
found in binary mode
Note that the leading zero does not display.
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Convert to decimal mode, then
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Divide by 2 raised to the number of digits (in
this case 7, including leading zero)
1
2
3
4
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In most cases it will not be exact
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Other way around
• Multiply fraction by 2 raised to the desired
number of digits in the fractional part. For
example
– .456  27 = 58.368
• Throw away the fractional part and represent the
whole number
– 58 111010
• But note that we specified 7 digits and the result
above uses only 6. Therefore we need to put in
the leading 0
– 0111010
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Fixed point
• If one has a set number of bits reserved for
representing the whole number part and
another set number of bits reserved for
representing the fractional part of a number,
then one is said to be using fixed point
representation.
– The point dividing whole number from fraction
has an unchanging (fixed) place in the number.
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Limits of the fixed point approach
• Suppose you use 4 bits for the whole
number part and 4 bits for the fractional part
(ignoring sign for now).
• The largest number would be 1111.1111 =
15.9375
• The smallest, non-zero number would be
0000.0001 = .0625
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Floating point representation
• Floating point representation allows one to
represent a wider range of numbers using
the same number of bits.
• It is like scientific notation.
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Scientific notation
• Used to represent very large and very small
numbers.
– Ex. Avogadro’s number
•  6.0221367  1023 particles
•  602213670000000000000000
– Ex. Fundamental charge e
•  1.60217733  10-19 C
•  0.000000000000000000160217733 C
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Scientific notation: all of these are the
same number
•
•
•
•
•
•
12345.6789
= 1234.56789  100
1234.56789  10 = 1234.56789  101
123.456789  100 =123.456789  102
12.3456789  103
1.23456789  104
Rule: Shift the point to the left and
increment the power of ten.
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Small numbers
•
•
•
•
•
•
•
•
0.000001234
0.00001234  10-1
0.0001234  10-2
0.001234  10-3
0.01234  10-4
0.1234  10-5
1.234  10-6
Rule: shift point to the right and decrement the
power.
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Floating Point Rules
• We’ll use a set of rules that are close but not quite
the same as the IEEE 754 standards for floating
point representation.
• Starting with the fixed point binary representation,
shift the point and increase the power (of 2 now
that we’re in binary).
• Shift so that the number has no whole number
part and also so that the first fractional bit (the
half’s place) has a 1.
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Floats
• SHIFT expression so it is just under 1 and
keep track of the number of shifts
• 1100010.1001100110011001
• .11000101001100110011001  27
• Express the number of shifts in binary
• .11000101001100110011001  200000111
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We’re not done yet so this
exponent will change.
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Mantissa and Exponent and Sign
•
•
•
•
•
.11000101001100110011001  200000111
(Significand) Mantissa
.11000101001100110011001  200000111
Exponent
The number may be negative, so there a bit
(the sign bit) reserved to indicate whether
the number is positive or negative
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Small numbers
• 0.000010101110
• 0.10101110  2-4
• The power (a.k.a. the exponent) could be negative
so we have to be able to deal with that.
• Floating point numbers use a procedure known as
biasing to handle the negative exponent problem.
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Biasing
• Actually the exponent is not represented as shown
on the previously.
• There were 8 bits used to represent the exponent
on the previous slide, that means there are 256
numbers that could be represented.
• Since the exponent could be negative (to represent
numbers less than 1), we choose half of the range
to be positive and half to be negative , i.e. -128 to
127.
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Biasing (Cont.)
• In biasing, one does not use 2’s
complement or a sign bit.
• Instead one adds a bias (equal to the
magnitude of the most negative number) to
the exponents and represents the result of
that addition.
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Biasing (Cont.)
• With 8 bits, the bias is 128 (= 27 that is 2 raised to
the number of bits used for the exponent minus
one).
• In our previous example, we had to shift 7 times to
the left, corresponding to an exponent of +7.
• We add that shift to the bias 128+7=135.
• That is the number we put in the exponent portion:
135  10000111.
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Big floats
• Assume we use 8 bits, 4 for the mantissa and 4 for
the exponent (neglecting sign). What is the largest
float?
• Mantissa: 1111 Exponent 1111
• 0.9375  27
• =120
• (Compare this to the largest fixed-point number
using the same amount of space 15.9375)
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Small floats
• Assume we use 8 bits, 4 for the mantissa and 4 for
the exponent (neglecting sign). What is the
smallest float?
• Mantissa: 1000 Exponent 0000
• 0.5  2-8
• = 0.001953125
• (Compare this to the smallest fixed-point number
using the same amount of space .0625)
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Adding Floats
• Consider adding the following numbers
expressed in scientific notation
3.456789  103
1.212121  10-2
• The first step is to re-express the number
with the smaller magnitude so that it has the
same exponent as the other number.
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Adding Floats (Cont.)
•
•
•
•
•
•
•
1.212121  10-2
0.1212121  10-1
0.01212121  100
0.001212121  101
0.0001212121  102
0.00001212121  103
The number was shifted 5 times (3-(-2)).
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Adding Floats (Cont.)
• When the exponents are equal the mantissas
can be added.
3.456789  103
0.00001212121  103
• =3.45680112121  103
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Rounding
• In a computer there are a finite number of
bits used to represent a number.
• When the smaller floating-point number is
shifted to make the exponents equal, some
of the less significant bits are lost.
• This loss of information (precision) is
known as rounding.
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One more fine point about
floating-point representation
• As discussed so far, the mantissa
(significand) always starts with a 1.
• When storage was expensive, designers
opted not to represent this bit, since it is
always 1.
• It had to be inserted for various operations
on the number (adding, multiplying, etc.),
but it did not have to be stored.
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Still another fine point
• When we assume that the mantissa must
start with a 1, we lose 0.
• Zero is too important a number to lose, so
we interpret the mantissa of all zeros and
exponent of all zeros as zero
– Even though ordinarily we would assume the
mantissa started with a one that we didn’t store.
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Yet another fine point
• In the IEEE 754 format for floats, you bias
by one less (127) and reserve the exponents
00000000 and 11111111 for special
purposes.
• One of these special purposes is “Not a
number” (NaN) which is the floating point
version of overflow.
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An example
• Represent -9087.8735 as a float using 23
bits for the mantissa, 8 for the exponent and
one for the sign.
• Convert the whole number magnitude 9087
to binary: 10 0011 0111 1111
• That uses up 14 of the 23 bits for the
mantissa, leaving 9 for the fractional part.
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An example (Cont.)
• Multiply the fractional part by 29 and
convert whole number part of that to binary,
make sure in uses 9 bits (add leading 0’s if
it doesn’t).
• .8735  29 = 447.232
• 447  110111111
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An example (Cont.)
•
•
•
•
•
10001101111111.110111111
.10001101111111110111111  214
Mantissa 10001101111111110111111
Exponent 14+128=142  10001110
Sign bit 1 (because number was negative)
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Example 2
• 0.0076534
• No whole number part. Begin by using all
23 mantissa bits for the fractional part.
• 0.0076534  223 = 64201.3724672
• 64201  1111101011001001
• Only uses 16 places, means that so far
number starts with 7 zeros. But float
mantissas are supposed to start with 1.
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Example 2 (Cont.)
23+7
•
•
•
•
•
0.0076534  230 = 8217775.6758016
821775  11111010110010010101111
Above is mantissa
Exponent 128 – 7 = 121  01111001
Sign bit 0 (positive number)
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References
• Computer Architecture, Nicholas Carter
• Computer Systems: Organization and
Architecture, John Carpinelli
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