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William Stallings
Computer Organization
and Architecture
Chapter 9
Computer Arithmetic
Rev. 3 (2005-06) by Enrico Nardelli
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Arithmetic & Logic Unit
• Does the calculations
• Almost everything else in the computer is there
to service this unit
• Handles integers
• May handle floating point (real) numbers
Rev. 3 (2005-06) by Enrico Nardelli
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ALU Inputs and Outputs
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Reminder
• Let A=an-1an-2…a1a0 be a binary number
• Its (decimal) value is:
n 1
A   2 ai
i
i 0
Example:
10010 = 24 + 21 = 16 + 2 = 18
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Conversion of integer to binary
• Represent (111)10 in pure binary
• Repeat division by 2
111:2= 55 remainder 1
55:2= 27 remainder 1
27:2= 13 remainder 1
13:2= 6 remainder 1
6:2= 3 remainder 0
3:2= 1 remainder 1
1:2= 0 remainder 1
• RESULT: 1101111
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Integer Representation
• Only have 0 & 1 to represent everything
• Positive numbers stored in binary
 e.g. 41=00101001
 Binary digit = Bit
• No minus sign
• Negative integer representations:
 Sign-Magnitude
 Two’s complement
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Sign-Magnitude
• Left most bit is sign bit
• 0 means positive
• 1 means negative
+18 = 00010010
-18 = 10010010
• Problems
 Need to consider both sign and magnitude in
arithmetic
 Two representations of zero (+0 and -0)
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A better representation:
Two’s Complement
• Let A=an-1an-2…a1a0 be a n-length binary string
• It’s value is in Two’s Complement representation is:
A  2
n 1
n2
an 1   2 ai
i
i 0
if an-1=0 then A is a non-negative number
if an-1=1 then A is a negative number
Rev. 3 (2005-06) by Enrico Nardelli
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Two’s Complement: examples
-128 64 32 16
0
1
0
0
8
4
2
1
0
1
1
0
+4
+2
8
4
2
1
0
0
1
1
+2
+1
+64
-128 64 32 16
1
0
0
0
-128
-128 64 32 16
1
-128
0
0
0
= 70
8
4
2
1
1
0
0
0
+8
Rev. 3 (2005-06) by Enrico Nardelli
= -125
= -120
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Benefits
• Only one representation of zero
+0
+1
+2
+3
+4
+5
+6
+7
=
=
=
=
=
=
=
=
0000
0001
0010
0011
0100
0101
0110
0111
-1
-2
-3
-4
-5
-6
-7
-8
=
=
=
=
=
=
=
=
1111
1110
1101
1100
1011
1010
1001
1000
• Arithmetic works easily (see later)
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Geometric Depiction of Two’s
Complement Integers
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Range of Numbers
• 8 bit 2’s complement
+127 = 01111111 = 27 -1
-128 = 10000000 = -27
• 16 bit 2’s complement
+32767 = 01111111 11111111 = 215 - 1
-32768 = 10000000 00000000 = -215
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2’s Complement representation for
negative numbers
•
To represent a negative number using the
“two’s complement” technique:
1. First decide how many bits are used for
representation
2. Then write the modulo of the negative number
3. Then, change each 0 in 1, each 1 in 0 (Boolean
Complement or “one’s complement”)
4. Finally, add 1 (as the result of Step 3 was a pure
binary number)
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Examples
• E.g.: how to represent -3 with 4 bits:
 Start from +3 = 0011
 Boolean complement gives
 Add 1 to LSB gives -3
• Represent -20 with 8 bits:
 Start from +20 =
 Bolean complement gives
 Add 1
1100
1101
00010100
11101011
11101100
• Negation works in the same way, e.g. negation of -3 is
obtained by the “two’s complement” of -3:
 Representation of -3
 Boolean complement gives
 Add 1 to LSB gives -(-3)=+3
1101
0010
0011
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Negation Special Case 1
•
•
•
•
•
•
0=
0000
Bitwise NOT
1111 (Boolean complement)
Add 1 to LSB
+1
Result
1 0000
Carry is ignored, so:
- 0 = 0 OK !
Rev. 3 (2005-06) by Enrico Nardelli
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Negation Special Case 2
•
•
•
•
•
•
•
•
-8 =
1000
Bitwise NOT
0111 (Boolean complement)
Add 1 to LSB
+1
Result
1000
So:
-(-8) = -8 WRONG !
Monitor MSB (sign bit)
If it does not change during negation there is a
mistake (but for 0!)
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Conversion Between Lengths
• Positive number: pack with leading zeroes
+18 =
00010010
+18 = 00000000 00010010
• Negative number: pack with leading ones
-18 =
11101110
-18 = 11111111 10010010
• i.e. pack with MSB (sign bit)
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Addition and Subtraction
• Addition: standard
• Overflow: when the result needs more bits to be
represented
• Monitor MSB: if it changes there may be an overflow
 When Pos + Pos or Neg + Neg the sign bit should not change: if
it does there is an overflow
• Subtraction: take two’s complement of subtrahend and
add to minuend
 i.e.:
a - b = a + (-b)
• So we only need addition and complement circuits
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Addition: examples
1001 + -7
0101 = 5
1110
-2
1100 + -4
0100 = 4
0
1 0000
1100 + -4
1111 = -1
-5
1 1011
0011 +
0100 =
0111
0101 +
0100 =
1001
1001 + -7
1010 = -6
overflow
1 0011
3
4
7
5
4
overflow
Rev. 3 (2005-06) by Enrico Nardelli
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Hardware for Addition and
Subtraction
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Multiplication
• Complex
• Work out partial product
for each digit
• Take care with place
value (column)
• Add partial products
1
1
0
1
1
X
1
1
0
1
=
1
0
1
1
+
0
0
0
0
-
+
1
0
1
1
-
-
+
1
0
1
1
-
-
-
=
0
0
0
1
1
1
1
Rev. 3 (2005-06) by Enrico Nardelli
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Multiplication Example
•
•
•
•
Multiplicand (11 dec)
Multiplier
(13 dec)
Sum partial products
Note: if multiplier bit is 1
copy multiplicand (place
value) otherwise put all
zeroes
• Product (143 dec)
• Note: need double length
1 0 1 1 X
1 1 0 1 =
1 0 1 1 +
0 0 0 0 - =
0 1 0 1 1 +
1 0 1 1 -
- =
1 1 0 1 1 1 +
1 0 1 1 -
-
- =
1 0 0 0 1 1 1 1
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Flowchart for Unsigned Binary
Multiplication
 A stores the most significant bits of the
result
 C is the carry bit for A
 Q initially stores the multiplier but at the
end stores the less significant bits of the
result





 C, A, Q are right shifted as a whole

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Execution of Example
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Unsigned Binary Multiplication
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Multiplying Negative Numbers
• Previous approach does not work!
• Solution 1
 Convert to positive if required
 Multiply as above
 If signs were different, negate answer
• Solution 2
 Booth’s algorithm
• Uses two’s complement representation
• More efficient
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Real Numbers
• …very informally, numbers with the
fractional point
• Actually, only finite precision real numbers
• we need to code the binary point
• we need to code the binary point position
• Solution: floating point numbers
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Reminder
• Let A=an-1an-2…a1a0 ,a-1,a-2,…,a-m be a binary number
• Its (decimal) value is:
n 1
A   2 ai 
i
i 0
1
2 a
j
j  m
j
Example:
1001,1011 = 23 + 20 +2-1 + 2-3 + 2-4 =9,6875
2-1
2-2
2-3
2-4
2-5
0,5
0,250
0,125
0,0625
0,03125
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Decimal to binary conversion (1)
• Represent (11,6875)10 in pure binary
• Integer part: iterate division taking the remainders
• Binary representation for (11)10 ?
11:2 = 5 remainder 1
5:2 = 2 remainder 1
2:2 = 1 remainder 0
1:2 = 0 remainder 1
Least Significant Bit
Most Significant Bit
• Result: 1011
• N.B.: Result bits are produced in reverse order
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Decimal to binary conversion (2)
• Fractional part: iterate multiplication taking the integer
parts
• Binary representation for 0,6875 ?
0,6875 x 2 = 1,375
0,375 x 2 = 0,75
0,750 x 2 = 1,5
0,5
x 2 = 1,0
take 1
take 0
take 1
take 1
MSB
LSB
• Result: 0,1011
• N.B.: Result bits here are produced MSB to LSB
• the procedure sometimes does not converge
 stop when the desired precision is reached
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Real Numbers
• Where is the binary point?
 Fixed?
• Very limited
 Moving?
• How do you show where it is?
• Example:
976.000.000.000.000
= 9,76 * 1014
0,0000000000000976
= 9,76 * 10-14
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Sign bit
Floating Point
Biased
Exponent
Significand or Mantissa
• Represents the value
+/- .<significand> * <base><exponent>
• “Floating” refers to the fact that the true
position of the point “floats” according to the
value of the exponent
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Normalization
(10,0101)2
= 10010,1 * 2-3
= 1001010 * 2-5
= 0,00100101 * 24
• Floating Point numbers are usually normalized
 exponent is adjusted so that there is a single bit equal to 1
before the binary point
• Similar to scientific notation for decimal numbers where
numbers are normalized to give a single digit before the
decimal point
3123 = 3.123 x 103
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Normalization
• A normalized number ( 0) has the following form:
+/- 1,bbb…b * 2 +/- E
where
b  {0,1}
• The base (or radix) for the exponent is suppose to be 2
(so it is not represented)
• Since MSB of the mantissa is always 1 there is no need
to represent it
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Representation of Floating Point
(IEEE Standard)
• Mantissa uses 23 bits to store a 24 bits number in the
interval [1,2)
• Sign is stored in the first bit
• Exponent value is represented in excess or biased
notation with 8 bits
• Excess (bias) 127 means




8 bit exponent field
Nominal exponent value has range 0-255
Subtract 127 (= 28-1-1) to get correct exponent value
Real range of exponent value is -127 to +128
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excess (bias) notation
• Two parameters are
specified:
 the number of bits n
 the bias value K
(usually, K=2n -1 -1)
• the string consisting
of all 0s represents
the value –K
• the string consisting
of all 1s represents
the value –K + 2n-1
n = 4 K=7
0000 = -7
0001 = -6
0010 = -5
0011 = -4
0100 = -3
0101 = -2
0110 = -1
0111 = 0
Rev. 3 (2005-06) by Enrico Nardelli
1000 =
1001 =
1010 =
1011 =
1100 =
1101 =
1110 =
1111 =
1
2
3
4
5
6
7
8
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Floating Point Examples
147 = 127+20
107 = 127-20
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Expressible Numbers
-(2-2-23)*2128
-2-127
2-127
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(2-2-23)*2128
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density
• Precision decreases with the increase of modulo
(representation is denser for smaller numbers)
• How many numbers can we represent in the interval
[2,4)?
• And in the range [4,8)?
• they are the same: 223
Note: Finite precision: (10.0/3.0)*3.0 may be different from 10.0
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Special cases
• Some strings are interpreted as special strings
• Two representations for zero (posit. and negat.)
 +/- 0 = 0/1 00000000 00000000000000000000000
• Able to represent infinity (posit. and negat.)
 +/-  = 0/1 11111111 00000000000000000000000
• Non-normalized numbers:
 0/1 00000000 (string different to all 0s)
• NAN = Not A Number
 Result of an operation which has no solution
 Propagated as NAN through subsequent operations
 representation: 0/1 11111111 (string different to all 0s)
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FP Arithmetic: addition and
subtraction
•
•
•
•
Check for zero
Align significands (adjusting exponents)
Add or subtract significands
Normalize result
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FP Arithmetic: multiplication
and division
•
•
•
•
•
•
Check for zero
Add/subtract exponents
Multiply/divide significands (watch sign)
Normalize
Round
All intermediate results should be in double
length storage
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Hexadecimal representation
• A compact representation for binary
numbers
• A byte (8 bit) is represented by 2
hex values
• Examples:
0011 1100 => 3C
0111 1111 1100 1001 -> 7FC9
• Used for numbers and code
Rev. 3 (2005-06) by Enrico Nardelli
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
9
A (10)
B (11)
C (12)
D (13)
E (14)
F (15)
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