Microsoft PowerPoint – CSE220 Unit02 Number Systems.pptx
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1Kevin McDonnell Stony Brook University – CSE 220
CSE 220:
Systems Fundamentals I
Unit 2:
Number Systems
2Kevin McDonnell Stony Brook University – CSE 220
Digital Abstraction
• Most physical variables are continuous
• Temperature
• Voltage on a wire
• Frequency of an oscillation
• Position of a mass
• Although voltage, charge and other electrical quantities are
continuous in nature, modern computers are all digital and
work with discrete values
• The continuous nature of electricity is “abstracted away”
and we consider only “high” and “low” voltage, or the
“presence” and “absence” of electric charge: i.e., 1s and 0s
(bits: binary digits)
3Kevin McDonnell Stony Brook University – CSE 220
Binary Digits
• So, all data is ultimately represented in a computer in
terms of binary digits
• Each bit is either a 0 or a 1
• Groups of bits represent larger values
• 1101 1110 1010 1101 1011 1110 1110 1111
• We usually write spaces between groups of four or eight
bits, depending on the situation. More on this soon.
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Positional Notation
• The scheme we use in modern times for representing
numbers is called positional notation
• The position of a digit determines how much it contributes
to the number’s value
• With decimal (base-10 or radix-10), the place-values are
powers of 10:
…, 103, 102, 101, 100, 10-1, 10-2, 10-3, …
…, 1000s, 100s, 10s, 1s, ⁄ s, ⁄ s, ⁄ s, …
• 642.15really means (6 102) + (4 101) + (2 100) +
(1 10-1) + (5 10-2)
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Positional Notation
• More generally, in base-10 notation, the sequence of digits
… stands for the polynomial expansion
× 10 + × 10 + ⋯ + × 10 +
× 10 + × 10
• We can generalize this to arbitrary bases. In radix-k we use
k distinct symbols (digits), and the place-values are powers
of k.
• Radix-2 (binary) notation example:
10101 = 1 × 2 + 0 × 2 + 1 × 2 + 0 × 2 + 1 × 2
= 16 + 4 + 1
= 21
• When working with multiple radixes, always include a
subscript to identify the radix!
6Kevin McDonnell Stony Brook University – CSE 220
Positional Notation
• In some circumstances it’s more natural to write numbers
in base 8, called octal, or base 16, called hexadecimal
• With octal there are 8 digits: 0, 1, 2, 3, 4, 5, 6, 7 with place-
values that are powers of 8:
…, 83, 82, 81, 80, 8-1, 8-2, 8-3, …
…, 256s, 64s, 8s, 1s, ⁄ s, ⁄ s, ⁄ s, …
• 376 = 3 × 8 + 7 × 8 + (6 × 8 )
= 192 + 56 + 6
= 254
• With hexadecimal we should have 16 digits: 0, 1, 2, 3, 4, 5,
6, 7, 8, 9, A, B, C, D, E, F, where the letters A through F
represent ten through fifteen, respectively
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Positional Notation
• Radix-16 (hexadecimal) notation example:
3E0 = 3 × 16 + 14 × 16 + (0 × 16 )
= 768 + 224 + 0
= 992
• Radix-k fractions involve negative powers of k
10.011 = 1 × 2 + 0 × 2 + 0 × 2 +
1 × 2 + 1 × 2
= 2 + 0.25 + 0.125
= 2.375
• Numbers with terminating decimal representations might
not have terminating representations in other radixes.
• For example: 0.2 = 0. 0011
8Kevin McDonnell Stony Brook University – CSE 220
Binary, Octal and Hexadecimal
• Note that the digits “10”
represent the base of a
number in its own base
• 10 is two
• 10 is eight
• 10 is sixteen
• 10 is ten
• Why does it work out like
this?
Binary Octal Hex Decimal
0 0 0 0
1 1 1 1
10 2 2 2
11 3 3 3
100 4 4 4
101 5 5 5
110 6 6 6
111 7 7 7
1000 10 8 8
1001 11 9 9
1010 12 A 10
1011 13 B 11
1100 14 C 12
1101 15 D 13
1110 16 E 14
1111 17 F 15
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9Kevin McDonnell Stony Brook University – CSE 220
Binary Decimal Conversion
• Algorithm: Start with initial value of 0. Process bits from
left-to-right order, i.e., from most significant bit (msb) to
least significant bit (lsb)
• Double the value from previous step.
• Add the next bit value.
10Kevin McDonnell Stony Brook University – CSE 220
Binary Decimal Example
• Convert 1011100 to decimal (going left to right)
1011100
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1 × 2 + 0 = 2
2 × 2 + 1 = 5
5 × 2 + 1 = 11
11 × 2 + 1 = 23
23 × 2 + 0 = 46
46 × 2 + 0 = 92
• This algorithm will work for any radix. Just multiply by the
radix after each step.
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Binary Decimal Example
• Convert 101011101 to decimal
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Octal Decimal Example
• Convert 417 to decimal
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Decimal Binary Conversion
• Algorithm: repeatedly divide the decimal representation by
2, writing the remainders in right-to-left order, i.e., from
least significant bit (lsb) to most significant bit (msb)
• Continue dividing until the quotient is 0
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Decimal Binary Example
• Convert 12310 to binary
• 123 / 2 = 61 rem. 1
• 61 / 2 = 30 rem. 1
• 30 / 2 = 15 rem. 0
• 15 / 2 = 7 rem. 1
• 7 / 2 = 3 rem. 1
• 3 / 2 = 1 rem. 1
• 1 / 2 = 0 rem. 1
• Answer: 11110112
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Decimal Binary Example
• Convert 1528 to binary
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Decimal Hexadecimal Example
• The decimal-to-hexadecimal conversion works largely in
the same way, but with division by 16
• 3241 / 16= 202 rem. 9
• 202 / 16 = 12 rem. 10
• 12 / 16 = 0rem. 12
• Answer: CA916
• This algorithm will also work for any radix. Just divide by
the radix after each step.
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Decimal Octal Example
• Convert 1528 to octal
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Decimal Fractions Binary
• Algorithm: generate the bits in left-to-right order, starting
from the radix point:
• Multiply the decimal value by 2. If the product is greater
than 1, the next bit is 1. Otherwise, the next bit is 0.
• Drop the integer part to get a value less than 1.
• Continue until 0 is reached (a terminating expansion) or
a pattern of digits repeats (a non-terminating expansion)
• The resulting representation is called fixed-point format
19Kevin McDonnell Stony Brook University – CSE 220
Decimal Frac. Binary Example
• Convert 0.4 to binary
• 0.4 × 2 = 0.80.8 < 1, so write a 0 ≅ 0.0• 0.8 × 2 = 1.61.6 ≥ 1, so write a 1 ≅ 0.01• Drop the integer part• 0.6 × 2 = 1.21.2 ≥ 1, so write a 1 ≅ 0.011• Drop the integer part• 0.2 × 2 = 0.4 0.4 < 1, so write a 0 ≅ 0.0110• Since we arrived at a decimal fraction we have already seen, the pattern will repeat• Final answer: 0. 011020Kevin McDonnell Stony Brook University – CSE 220Decimal Frac. Binary Example• Convert 13.85 to binary621Kevin McDonnell Stony Brook University – CSE 220Bin ↔ Oct ↔ Hex Conversion• Because 8 and 16 are powers of 2, converting between bases 2 and 8, and between bases 2 and 16 is very simple• Binary Octal• Working right-to-left, take bits in groups of 3, converting the groups into octal digits (Why 3? Because 2 = 8. ) • Example: 10110101→ 10 110 101 → 265• Binary Hexadecimal• Working right-to-left, take bits in groups of 4 (a nibble), converting the groups into hexadecimal digits (2= 16)• Example: 10110111101→ 101 1011 1101 → 5BD• Two nibbles = eight bits = one byte22Kevin McDonnell Stony Brook University – CSE 220Bin ↔ Oct ↔ Hex Conversion• Octal Binary• Working left-to-right, replace each octal digit with its 3-bit binary equivalent• Example: 250→ 010 101 000 → 10101000• Hexadecimal Binary• Working left-to-right, replace each hexadecimal digit with its 4-bit binary equivalent• Example: 3FE5→ 0011 1111 1110 0101 →11111111100101• Conversion between octal and hexadecimal is not so straightforward. It’s easiest to convert the given representation into binary and then into the desired base.23Kevin McDonnell Stony Brook University – CSE 220Bin ↔ Oct ↔ Hex Conversion• Convert 7BA3. BC4 to octal• Convert 2713 to hexadecimal24Kevin McDonnell Stony Brook University – CSE 220Base ↔ Base • General algorithm for converting from base 2 to 2• Write out the binary equivalent of the base 2 number and, going right-to-left, form groups of N bits• Convert 2713 to base 4• Because 4 = 2 , we will take bits in pairs (right-to-left)• General algorithm for converting from base 2 to 2• Write out the binary equivalent of the base 2 number and, going right-to-left, form groups of N bits725Kevin McDonnell Stony Brook University – CSE 220General Base Conversions• What if we want to convert between two bases other than the ones we’ve studied?• Generally it’s easiest to convert the given representation into decimal, and then from decimal into the desired base• Example: convert 215 to base 5• 215 = 2 × 7 + 1 × 7 + 5 × 7= 98 + 7 + 5= 110• 110 / 5 = 22 rem. 0• 22 / 5 = 4 rem. 2• 4 / 5 = 0 rem. 4• Answer: 215 = 42026Kevin McDonnell Stony Brook University – CSE 220Integer Encodings: Unsigned• Now we’ll start to see how integers are encoded in hardware• In unsigned integer encodings, the numbers 0, 1, 2, …, 2 − 1 are typically encoded as follows:0 → 000 … 001 → 000 … 012 → 000 … 103 → 000 … 11…2 − 1 → 111 … 1127Kevin McDonnell Stony Brook University – CSE 220Integer Encodings: Unsigned• Binary arithmetic with unsigned integers (particularly, addition), is done in the usual way• With decimal addition we can have carried values:Carry values:11 7410+8910Sum: 16310• The same thing can happen with binary additionCarry values: 1 1 1 0 1 0 0 1 0 1 02+ 0 1 0 1 1 0 0 12Sum: 1 0 1 0 0 0 1 1228Kevin McDonnell Stony Brook University – CSE 220Binary Addition Example• Add the following two 8-bit binary numbers:1 1 0 1 0 1 1 12+ 0 1 0 1 0 0 0 12• We had an overflow. • In fixed-precision integer arithmetic, it is possible for an arithmetic operation, such as addition, to result in an overflow. The leftmost carry value is dropped, resulting in an incorrect sum.• The programmer must be aware of the range of representable values to avoid unintentional overflow829Kevin McDonnell Stony Brook University – CSE 220Integer Encodings: Signed• There are several different signed integer encodings which permit the representation of both positive and negative numbers.• The hardware designer chooses between the encodings to make the arithmetic hardware simpler or more efficient for certain operations.• Sometimes the programmer needs to be aware of what encoding is being used:• To format numbers for input or output.• To understand the range of representable values and the conditions under which overflow can occur.30Kevin McDonnell Stony Brook University – CSE 220Sign/Magnitude Numbers• In N-bit sign/magnitude encoding, the most significant bit (leftmost bit) is used as a sign bit (0 = “positive”, 1 = “negative”), and the remaining − 1 bits represent the magnitude (absolute value) of the number, as in the unsigned scheme.• Range: −2 + 1, 2 − 1• Example (8-bit precision): • +75 ⇒ 01001011• −15 ⇒ 10001111• Problems with sign/magnitude encoding:• Two encodings for zero: +0 and −0• Subtraction is somewhat complicated31Kevin McDonnell Stony Brook University – CSE 220One’s Complement Encoding• The N-bit one’s complement encoding represents integers in the range − 2 − 1 , 2 − 1 as follows:0 → 000 … 00 −0 → 111 … 111 → 000 … 01 −1 → 111 … 102 → 000 … 10 −2 → 111 … 013 → 000 … 11 −3 → 111 … 00… …2 − 1 → 011 … 11 − 2 − 1 → 100 … 00• To obtain the negative of a value, complement (“flip”) all the bits:• change all 0s to 1s and all 1s to 0s32Kevin McDonnell Stony Brook University – CSE 220One’s Complement Encoding• Addition may require an “end around carry”:0 0 0 0 0 1 0 1 5+ 1 1 1 1 1 1 0 0 -31 0 0 0 0 0 0 0 1+ 10 0 0 0 0 0 1 0 2933Kevin McDonnell Stony Brook University – CSE 220One’s Complement Example• Perform the computation −28 − 37 in 8-bit one’s complement and convert the result to decimal.34Kevin McDonnell Stony Brook University – CSE 220One’s Complement Issue• The presence of two representations for zero adds complexity to a circuit that implements addition• So we can use a different encoding called two’s complement that avoids this problem35Kevin McDonnell Stony Brook University – CSE 220Two’s Complement Encoding• The N-bit two’s complement encoding represents integers in the range −2 , 2 − 1 as follows:0 → 000 … 001 → 000 … 01 −1 → 111 … 112 → 000 … 10 −2 → 111 … 103 → 000 … 11 −3 → 111 … 01… …2 − 1 → 011 … 11 −2→ 100 … 0036Kevin McDonnell Stony Brook University – CSE 220Two’s Complement Encoding• To negate a number in two’s complement encoding, use one of the following (equivalent) rules:1. Complement all the bits and increment the result.Example: negate 300000011 → 11111100 → 111111012. Complement all the bits to the left of the rightmost 1.Example: negate 900001001 → 11110111• The value 2 has no representation in N-bit two’s complement notation, but −2 does.1037Kevin McDonnell Stony Brook University – CSE 220Two’s Complement• What is the 8-bit two’s complement representation of −99?• What is the decimal equivalent ofthe two’s complement number 11001010?38Kevin McDonnell Stony Brook University – CSE 220Two’s Complement of Zero• What happens if we take the two’s complement of the number 0 written as an 8-bit number?39Kevin McDonnell Stony Brook University – CSE 220The Two’s Complement Wheel• Unlike mathematical integers, which inhabit a number line, in computer arithmetic the numbers lie on a circle.• An overflow occurs when crossing the boundary between the greatest representable positive number and the least representable negative number.Image: users.dickinson.edu/~braught/courses/cs251f02/classes/notes07.html40Kevin McDonnell Stony Brook University – CSE 220The Two’s Complement Wheel1141Kevin McDonnell Stony Brook University – CSE 220Two’s Complement Encoding• The leftmost bit is the sign bit, which tells whether the number is positive (0) or negative (1).• The rightmost bit tells whether the number is even (0) or odd (1).• Multiplication by 2 is accomplished by a left shift(introduce 0 at right), dropping the msb:−3 → −611111101 → 11111010• Division by 2 is accomplished by a right arithmetic shift(replicate sign bit at left), dropping the lsb:−6 → −311111010 → 1111110142Kevin McDonnell Stony Brook University – CSE 220Overflow in Two’s Complement• In signed arithmetic using two’s complement encoding, there is a possibility of overflow, when the correct result is “too large” or “too small” to be represented.• Overflow cannot occur when adding numbers of opposite sign (or when subtracting numbers of the same sign).• Overflow can occur when adding numbers of the same sign (or subtracting numbers of opposite sign).0 1 1 1 1 1 1 0 126+ 0 0 0 0 0 0 1 1 +31 0 0 0 0 0 0 1 -127• The overflow can be detected by noticing that the sum has opposite sign from the addends.43Kevin McDonnell Stony Brook University – CSE 220Sign Extension• When a two’s complement number is extended to more bits, the sign bit must be copied into the most significant bit positions• This is called sign-extension• Examples: rewrite each of the following numbers (3 and − 3) in 8-bit two’s complement• 3 is 0011 00000011• −3 is 1101 1111110144Kevin McDonnell Stony Brook University – CSE 220Excess-k Encoding • Excess-k encoding is another signed integer encoding that is important due to its use in representing real numbers • In excess-k, also called bias-k, integer i is represented by the unsigned encoding of + . • For example, in excess-127:• 3 is represented as 3 + 127 = 130→ 10000010• −5 is represented as −5 + 127 = 122→ 01111010• We note that both positive and negative numbers are represented as unsigned values. In excess-127:• −127 is 00000000 and +128 is 11111111• The advantage of excess-k notation is that ordering of positive and negative numbers is preserved, which makes comparing two values (e.g., <, >) very straightforward
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Comparison of Integer Encodings
N
decimal
N
binary
−
sign/mag
−
1’s comp.
−
2’s comp.
−
bias-127
1 00000001 10000001 11111110 11111111 01111110
2 00000010 10000010 11111101 11111110 01111101
3 00000011 10000011 11111100 11111101 01111100
4 00000100 10000100 11111011 11111100 01111011
5 00000101 10000101 11111010 11111011 01111010
10 00001010 10001010 11110101 11110110 01110101
50 00110010 10110010 11001101 11001110 01001101
90 01011010 11011010 10100101 10100101 00100101
100 01100100 11100100 10011011 10011100 00011011
127 01111111 11111111 10000000 10000001 00000000
128 10000000 N/A N/A 10000000 N/A
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What About Real Numbers?
• We did some base conversions involving real numbers but
haven’t seen yet how they can be represented in a
computer
• A big disadvantage of the so-called fixed-point format or
fixed-precision encoding of such numbers is that they
have a very limited “dynamic range”
• This forma can’t represent very large numbers (e.g., 2 )
or very small numbers (e.g., 2 )
• These numbers are called “fixed point” because the
decimal point (or binary point) is fixed, which limits
accuracy
• On the other hand, they give exact answers (i.e., no
rounding errors) as long as there is no overflow
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Floating-Point Format
• Because fractions can have non-terminating
representations and/or might require many digits to be
represented exactly, usually real numbers can only be
approximately represented in a computer
• We have to tolerate a certain amount of
representational error
• The industry standard way used to approximate real
numbers is called floating-point format
• IEEE 754 floating-point standard
• In this scheme the binary point is allowed to “float” (i.e., be
repositioned) in order to give as accurate an
approximation as possible
48Kevin McDonnell Stony Brook University – CSE 220
IEEE 754 Floating-Point Standard
• The IEEE 754 standard species floating-point
representations of numbers and also arithmetic operations
on these representations
• IEEE 754 is essentially a form of scientific notation, but
written in binary: ±2 ×
• This format can be encoded using three fields: a sign bit
(s), an exponent (e) and a fraction (f ), sometimes called
the mantissa
• IEEE 754 single-precision format requires 32 bits and
provides about 7 decimal digits of accuracy
• IEEE 754 double-precision format requires 64 bits and
provides about 15 decimal digits of accuracy
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49Kevin McDonnell Stony Brook University – CSE 220
IEEE 754 Floating-Point Standard
• 32-bit version:
• 64-bit version:
• Sign bit: 0(positive) or 1 (negative)
• Exponent: stored in excess-127 for the 32-bit version
• Excess-1023 for the 64-bit version
• Fraction: contains the digits to the right of the binary point
• Normalized: the digit to the left of the point is always 1,
and is not represented, giving us one bit of precision “for
free”
1 bit 8 bits 23 bits
sign exponent fraction (mantissa)
1 bit 11 bits 52 bits
sign exponent fraction (mantissa)
single precision
double precision
50Kevin McDonnell Stony Brook University – CSE 220
IEEE 754 to Decimal
• Decimal value of a IEEE 754 floating point encoding is
given by the formula: −1 × 2 × 1 +
where:
• s is the sign bit (0/1).
• e is the decimal value of the exponent field
• bias is 127 for single-precision, 1023 for double-
precision.
• f is the decimal value of the fraction field (regarded as a
binary fraction)
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IEEE 754 to Decimal Example
• What decimal value has the following IEEE 754 encoding?
10111110011000000000000000000000
1 01111100 11000000000000000000000
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Decimal to IEEE 754 Example
• Encode 13.4 in 32-bit IEEE 754 floating-point format
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IEEE 754 Special Values
• The smallest 000…0 and largest 111…1 exponents are
reserved for the encoding of special values:
• Zero (two encodings):
• = 0 or 1, = 000 … 0, = 000 … 0
• Infinity:
• +∞: = 0, = 111 … 1, = 000 … 0
• −∞: = 1, = 111 … 1, = 000 … 0
• NaN (not a number):
• = 0 or 1, = 111 … 1, = non-zero
• Can result from division by zero, −1, log (−5), etc.
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IEEE 754 Format Summary
Property Single-Precision Double-Precision
Bits in Sign 1 1
Bits in Exponent 8 11
Bits in Fraction 23 52
Total Bits 32 64
Exponent Encoding excess-127 excess-1023
Exponent Range −126 to 127 −1022 to 1023
Decimal Range ≅ 10 to 10 ≅ 10 to 10
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Floating-Point Limitations
• There are ~2 different values that can be represented in
single-precision floating point.
• This is the same as the number of values that can be
represented using 32-bit integer encodings.
• Many values (even integers) do not have floating-point
representations:
• Examples: 33554431 and 0.33554431
• Try assigning these to a float variable in Java and then
printing them out.
• Caution: Results of floating point calculations are not exact.
• Never use floating point when exact results are essential or
in equality checks, such as in conditional statements
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