2.4 Floating-Point Representation & IEEE 754

Theory: Floating-Point Representation & IEEE 754

Integers alone are not enough for scientific or business computing. This section shows how computers store very large, very small and fractional numbers by borrowing the idea of scientific notation, first with the lecture’s simple 14-bit model, then with the real-world IEEE 754 standard (Null & Lobur, Ch. 2).

Why Floating Point?

With 16-bit two’s complement integers the largest value is $+32{,}767$ . Two everyday needs break this immediately:

Large values: how do we store $32{,}000{,}000$ ?
Fractions: how do we store $9.625$ ?

The answer is floating-point representation, which is scientific notation in binary: $32{,}000{,}000 = 32 \times 10^6$ needs only the digits “32” and the exponent “6”.

Floating-point calculations can be done using only integer hardware, by software that stores the parts separately and manipulates them. This is called floating-point emulation (floating-point values are not stored as such; programs make it seem as if they are). Most of today’s computers instead have specialized floating-point hardware (an FPU), which is far faster.

The Three Parts of Scientific Notation

\underbrace{+}_{\text{sign}}\ \underbrace{1.25}_{\text{mantissa (significand)}} \times 10^{\overbrace{-1}^{\text{exponent}}}

Part	Meaning
Sign	positive or negative
Significand (mantissa)	the significant digits; its length sets the precision
Exponent	the power of the base; its size sets the range

The radix point “floats”: the same number can be written many ways ( $17 = 17.0 \times 10^0 = 1.7 \times 10^1 = 0.17 \times 10^2$ ), and the exponent records where the point really is. In binary, $17 = 10001_2 \times 2^0 = 1000.1_2 \times 2^1 = \cdots = 1.0001_2 \times 2^4$ .

Normalization

To make the representation unique and use the significand bits efficiently, numbers are normalized: the exponent is adjusted so there are no leading 0s, giving the form

1.\text{xxxxxxxx} \times 2^{\text{yyyy}}

For example $0.01 \times 2^{-1} = 0.1 \times 2^{-2} = 1.0 \times 2^{-3}$ .

Shifting rule: moving the binary point left by $k$ places adds $k$ to the exponent; moving it right by $k$ places subtracts $k$ . Example: $1001.101 = 1.001101 \times 2^3$ (point moved left 3).

Since a normalized nonzero binary number always starts with “1.”, that 1 need not be stored. Only the fraction bits after the point are kept; the leading 1 is implicit (a “hidden bit”), which gains one bit of precision for free.

The Lecture’s 14-bit Model

Field	Bits	Contents
Sign	1	0 = positive, 1 = negative
Exponent	5	power of 2
Significand	8	the 8 fraction bits after the implicit leading 1

13sign (1 bit)

12–8exponent (5 bits)

7–0significand (8 bits)

Step 1: the simple (unbiased) model

Store the exponent as a plain unsigned 5-bit number. For $17 = 1.0001 \times 2^4$ : sign 0, exponent $4 = 00100$ , fraction $0001$ padded to 8 bits $= 00010000$ :

17 \;\to\; 0\ \ 00100\ \ 00010000

Disadvantage: a 5-bit unsigned field only holds $0$ to $31$ , so there is no way to store a negative exponent, e.g. $1.0001 \times 2^{-4}$ . Every small number (between 0 and 1) is impossible.

Step 2: the biased (excess-16) model

To allow negative exponents, store the exponent plus a fixed bias. With 5 bits there are 32 values ( $0$ to $31$ ); choosing the bias near the middle, 16, gives an excess-16 representation:

\text{stored exponent} = \text{true exponent} + 16, \qquad \text{true exponent} = \text{stored} - 16

Stored exponent	True exponent
$0$ to $15$	negative ( $-16$ to $-1$ )
$16$	zero
$17$ to $31$	positive ( $+1$ to $+15$ )

Advantage: biased exponents are plain unsigned numbers, so comparing or sorting floating-point values is easier and faster: a larger true exponent always has a larger stored pattern. For example $-1 \to 15 = 01111$ and $-2 \to 14 = 01110$ ; since $01111 \gt 01110$ as unsigned numbers, $-1 \gt -2$ is preserved.

Mini-example

$17 = 1.0001 \times 2^4$ : stored exponent $16 + 4 = 20 = 10100$ , so $17 \to 0\ 10100\ 00010000$ .

$-1.0100 \times 2^{-9}$ : sign 1, stored exponent $16 - 9 = 7 = 00111$ , fraction $01000000$ , so it is stored as $1\ 00111\ 01000000$ .

Procedure: decimal → 14-bit (excess-16)

Sign bit: 0 if positive, 1 if negative. Work with the absolute value from now on.
Convert to binary (integer part by division, fraction part by multiplication).
Normalize to $1.f \times 2^e$ .
Exponent field $= e + 16$ in 5-bit binary (must lie in $0$ to $31$ ).
Significand field = the bits of $f$ , padded with 0s on the right (or truncated) to 8 bits.

The IEEE 754 Standard

Set by the Institute of Electrical and Electronics Engineers (IEEE) in 1985, to make floating-point programs portable between processors and to encourage numerically oriented software. It uses the same ideas as the 14-bit model: sign, biased exponent and fraction with an implicit leading 1.

Format	Total	Sign	Exponent	Fraction	Bias	Special exponent
Single precision	32 bits	1	8	23	127	255 (all ones)
Double precision	64 bits	1	11	52	1023	2047 (all ones)

The value of a normal number is

(-1)^S \times (1.F)_2 \times 2^{\,E - \text{bias}}

where $S$ is the sign bit, $E$ the stored (biased) exponent and $F$ the stored fraction.

31sign (1 bit)

30–23exponent (8 bits)

22–0fraction (23 bits)

Single precision (32 bits)

63sign (1 bit)

62–52exponent (11 bits)

51–0fraction (52 bits)

Double precision (64 bits)

Special values (single precision; for double replace 255 by 2047):

Exponent $E$	Fraction $F$	Meaning
0	0	zero: $+0$ if $S = 0$ , $-0$ if $S = 1$ (two zeros)
1 to 254	anything	normal number $(-1)^S \times 1.F \times 2^{E-127}$
255	0	infinity: $+\infty$ or $-\infty$ depending on $S$
255	nonzero	NaN (“not a number”), flags an error, e.g. $0/0$ or $\sqrt{-1}$

Because zero has two representations ( $+0$ and $-0$ ) and because results are approximations, programmers should avoid testing a floating-point value for equality to zero (see section 2.5).

Procedure: decimal → IEEE single

Sign bit $S$ .
Convert $|x|$ to binary and normalize to $1.f \times 2^e$ .
$E = e + 127$ as an 8-bit unsigned number.
$F$ = bits of $f$ , padded to 23 bits.
Write $S\,E\,F$ as 32 bits and group into 4s for hex.

Procedure: IEEE single → decimal: expand the hex into 32 bits, split $1 | 8 | 23$ , check for special values, otherwise compute $(-1)^S \times 1.F \times 2^{E-127}$ .

Mini-example

$1.0$ : $S = 0$ , $1.0 = 1.0 \times 2^0$ , $E = 0 + 127 = 01111111$ , $F = 0$ : $0\ 01111111\ 000\cdots0 = 0011\,1111\,1000\,0000 \cdots = \text{3F800000}_{16}$ .

Worked Examples

Exercise 1

Store $17_{10}$ in the lecture’s 14-bit floating-point model (a) with an unsigned (unbiased) exponent and (b) with an excess-16 exponent. Why was the bias introduced?

Show solution ↓

Solution

Convert and normalize: $17 = 16 + 1 = 10001_2 = 1.0001_2 \times 2^4$ (point moved left 4 places).

Sign: positive → $0$
Fraction after the leading 1: $0001$ → padded to 8 bits: $00010000$

(a) Unbiased: exponent field $= 4 = 00100$ .

17 \;\to\; 0\ \ 00100\ \ 00010000

(b) Excess-16: exponent field $= 4 + 16 = 20 = 10100$ .

17 \;\to\; 0\ \ 10100\ \ 00010000

Check (b): exponent $20 - 16 = 4$ ; significand $1.0001_2 = 1.0625$ ; value $1.0625 \times 16 = 17$ ✓.

Why a bias? An unsigned exponent field cannot hold negative exponents, so numbers like $1.0001 \times 2^{-4}$ (anything smaller than 1) could not be stored. With excess-16, stored values $0$ to $15$ mean exponents $-16$ to $-1$ . The bias also keeps exponent fields unsigned, so magnitudes compare like ordinary binary numbers (fast sorting/comparison).

Exercise 2

Express each value in the 14-bit model with an excess-16 exponent:

(a) $9.625$ (b) $-0.375$ (c) $-1.0100_2 \times 2^{-9}$ (also give its decimal value)

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Solution

(a) $9 = 1001_2$ ; $0.625$ : $0.625 \times 2 = 1.25 \to 1$ , $0.25 \times 2 = 0.5 \to 0$ , $0.5 \times 2 = 1.0 \to 1$ , so $0.625 = 0.101_2$ .

9.625 = 1001.101_2 = 1.001101_2 \times 2^3

Sign $0$
Exponent $3 + 16 = 19 = 10011$
Fraction $001101$ → $00110100$

Answer: $0\ 10011\ 00110100$ . Check: $1.001101_2 = 1 + \tfrac18 + \tfrac1{16} + \tfrac1{64} = 1.203125$ , and $1.203125 \times 8 = 9.625$ ✓.

(b) $0.375$ : $0.375 \times 2 = 0.75 \to 0$ , $0.75 \times 2 = 1.5 \to 1$ , $0.5 \times 2 = 1.0 \to 1$ , so $0.375 = 0.011_2 = 1.1_2 \times 2^{-2}$ (point moved right 2).

Sign $1$
Exponent $-2 + 16 = 14 = 01110$
Fraction $1$ → $10000000$

Answer: $1\ 01110\ 10000000$ . Check: $-1.5 \times 2^{-2} = -0.375$ ✓.

(c) Already normalized.

Sign $1$
Exponent $-9 + 16 = 7 = 00111$
Fraction $0100$ → $01000000$

Answer: $1\ 00111\ 01000000$ . Decimal value: $1.0100_2 = 1.25$ , so $-1.25 \times 2^{-9} = -1.25 / 512 = -0.00244140625$ .

Exercise 3

Each pattern is in the 14-bit excess-16 model. Find its decimal value.

(a) $0\ 10010\ 10100000$ (b) $1\ 01110\ 10000000$ (c) $0\ 10001\ 10000000$ (d) $0\ 01100\ 10011001$

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Solution

Value $= (-1)^S \times 1.\text{(fraction)}_2 \times 2^{\text{exp} - 16}$ .

(a) $S = 0$ ; exponent $10010 = 18$ , true $18 - 16 = 2$ ; significand $1.101_2 = 1.625$ . $1.625 \times 2^2 = 6.5$ . (Or shift: $1.101 \times 2^2 = 110.1_2 = 6.5$ .) $+6.5$

(b) $S = 1$ ; exponent $01110 = 14$ , true $-2$ ; significand $1.1_2 = 1.5$ . $-1.5 \times 2^{-2} = -0.375$ . $-0.375$

(c) $S = 0$ ; exponent $10001 = 17$ , true $1$ ; significand $1.1_2 = 1.5$ . $1.5 \times 2 = 3$ . $+3$

(d) $S = 0$ ; exponent $01100 = 12$ , true $-4$ ; significand $1.10011001_2 = 1 + \tfrac{153}{256} = 1.59765625$ . $1.59765625 / 16 = 0.099853515625$ . This is the closest the model gets to $0.1$ (by truncation), a first taste of the precision limits discussed in section 2.5.

Exercise 4

In the excess-16 model:

(a) Give the stored exponent field for true exponents $-1$ , $-2$ , $5$ and $6$ , and show that comparing the stored fields as unsigned numbers gives the correct order.

(b) What range of true exponents can be stored? What is the largest positive value of the 14-bit model?

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Solution

(a) Stored $=$ true $+ 16$ :

True exponent	Stored (decimal)	Stored (binary)
$-2$	14	01110
$-1$	15	01111
$5$	21	10101
$6$	22	10110

As unsigned numbers $01110 \lt 01111 \lt 10101 \lt 10110$ , which is exactly $-2 \lt -1 \lt 5 \lt 6$ . With a two’s complement exponent instead, $-1$ would be $11111$ and would wrongly look larger than $6 = 00110$ .

(b) The 5-bit field holds $0$ to $31$ , so true exponents are $0 - 16 = -16$ up to $31 - 16 = +15$ .

Largest positive value: sign 0, largest exponent $11111$ ( $+15$ ), all fraction bits 1:

0\ 11111\ 11111111 = 1.11111111_2 \times 2^{15} = (2 - 2^{-8}) \times 2^{15} = 65536 - 128 = 65408

Any result larger than this overflows. (At the small end the true exponent cannot go below $-16$ ; anything needing a smaller exponent underflows.)

Exercise 5

Convert to IEEE 754 single precision. Give the 32 bits and the hexadecimal form.

(a) $-12.375$ (b) $0.15625$

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Solution

(a) $-12.375$

Sign: negative → $S = 1$ .
$12 = 1100_2$ ; $0.375 = 0.011_2$ (from $0.375 \to 0.75 \to 1.5 \to 1.0$ : digits 0, 1, 1). So $12.375 = 1100.011_2$ .
Normalize: $1100.011 = 1.100011 \times 2^3$ .
$E = 3 + 127 = 130 = 1000\,0010_2$ .
$F = 100011$ followed by 17 zeros (23 bits).

311S (1 bit)

30–2310000010E (8 bits)

22–010001100000000000000000F (23 bits)

Group into 4s: $1100\ 0001\ 0100\ 0110\ 0000\ 0000\ 0000\ 0000$

Hex: $\text{C1460000}_{16}$ .

Check: $-(1 + \tfrac12 + \tfrac1{32} + \tfrac1{64}) \times 2^{130 - 127} = -1.546875 \times 8 = -12.375$ ✓.

(b) $0.15625$

$S = 0$ .
$0.15625 \times 2 = 0.3125 \to 0$ ; $0.3125 \times 2 = 0.625 \to 0$ ; $0.625 \times 2 = 1.25 \to 1$ ; $0.25 \times 2 = 0.5 \to 0$ ; $0.5 \times 2 = 1.0 \to 1$ . So $0.15625 = 0.00101_2$ .
Normalize: $0.00101 = 1.01 \times 2^{-3}$ (point moved right 3).
$E = -3 + 127 = 124 = 0111\,1100_2$ .
$F = 01$ followed by 21 zeros.

310S (1 bit)

30–2301111100E (8 bits)

22–001000000000000000000000F (23 bits)

Group into 4s: $0011\ 1110\ 0010\ 0000\ 0000\ 0000\ 0000\ 0000$

Hex: $\text{3E200000}_{16}$ .

Check: $1.25 \times 2^{124 - 127} = 1.25 / 8 = 0.15625$ ✓.

Exercise 6

Interpret each IEEE 754 single-precision hex pattern.

(a) $\text{41C80000}$ (b) $\text{C0A00000}$ (c) $\text{7F800000}$ (d) $\text{FF800000}$ (e) $\text{7FC00000}$ (f) $\text{80000000}$

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Solution

Expand to binary and split $1 | 8 | 23$ .

(a) $\text{41C80000} = 0100\,0001\,1100\,1000\,0000\cdots$

$S = 0$
$E = 1000\,0011 = 131$ , so $e = 131 - 127 = 4$
$F = 1001\,000\cdots$ , significand $1.1001_2$

Value: $1.1001_2 \times 2^4 = 11001_2 = 25$ . $+25$

(b) $\text{C0A00000} = 1100\,0000\,1010\,0000\cdots$

$S = 1$
$E = 1000\,0001 = 129$ , $e = 2$
$F = 0100\cdots$ , significand $1.01_2 = 1.25$

Value: $-1.25 \times 4 = -5$ . $-5$

(c) $\text{7F800000} = 0\ 11111111\ 000\cdots0$ : $E = 255$ , $F = 0$ , $S = 0$ → $+\infty$ .

(d) $\text{FF800000} = 1\ 11111111\ 000\cdots0$ : $E = 255$ , $F = 0$ , $S = 1$ → $-\infty$ .

(e) $\text{7FC00000} = 0\ 11111111\ 1000\cdots0$ : $E = 255$ , $F \ne 0$ → NaN.

(f) $\text{80000000} = 1\ 00000000\ 000\cdots0$ : $E = 0$ , $F = 0$ , $S = 1$ → $-0$ (the second zero; $\text{00000000}$ is $+0$ ).

Exercise 7

(a) Write $1.0$ and $-2.5$ in IEEE 754 double precision (hex).

(b) Why does the double-precision format reach magnitudes of about $10^{308}$ ? What is the approximate largest single-precision value?

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Solution

(a) Double: 1 sign bit, 11 exponent bits with bias 1023, 52 fraction bits.

$1.0 = 1.0 \times 2^0$ : $S = 0$ , $E = 0 + 1023 = 011\,1111\,1111$ , $F = 0$ .

0 01111111111 0000...0000
= 0011 1111 1111 0000 ... 0000

Hex: $\text{3FF0000000000000}_{16}$ .

$-2.5 = -10.1_2 = -1.01_2 \times 2^1$ : $S = 1$ , $E = 1 + 1023 = 1024 = 100\,0000\,0000$ , $F = 01$ then 50 zeros.

1 10000000000 0100...0000
= 1100 0000 0000 0100 0000 ... 0000

Hex: $\text{C004000000000000}_{16}$ .

(b) With 11 exponent bits and bias 1023, true exponents go up to about $\pm 1023$ , so magnitudes reach roughly $2^{1024}$ . Converting to a power of 10, solve $2^{1024} = 10^x$ :

x = 1024 \log_{10} 2 \approx 1024 \times 0.30103 \approx 308.25

So doubles span roughly $10^{-308}$ to $10^{308}$ in magnitude. Anything beyond is overflow; anything nonzero but closer to 0 than about $10^{-308}$ is underflow.

Single precision (8-bit exponent, largest true exponent $254 - 127 = 127$ ): largest value $= (2 - 2^{-23}) \times 2^{127} \approx 3.4 \times 10^{38}$ ; the smallest normal positive value is $2^{-126} \approx 1.18 \times 10^{-38}$ .