Floating Point Numbers

Topics Covered

Fixed point Numbers

Representation of Floating Point NumbersIEEE 32-bit floating point number.

Floating point Arithmetic

Fixed Point Numbers

The binary (or decimal) point is assumed to be in a fixed position

Base 10 fixed point arithmetic:

7632135 763.21351794821 179.48219426956 942.6956

Fixed Point (Binary) Numbers

Example: Add 3.625 and 6.5

1. Convert the numbers to 8-bit form (4-bit int, 4-bit fraction):3.625 11.101 0011.10106.500 110.10 0110.1000

2. Consider the numbers having an imaginary binary point and added in the normal way:

00111010 + 01101000 = 10100010

3. The integer part of the result is converted to 10, and the fractional part is interpreted as .125. Therefore, the result is 10.125.

Problem with Fixed Point (Binary) Numbers

Some systems require a large range of numbers:

1. Mass of sun: 1990000000000000000000000000000000 gramsRequires about 14 bytes

2. Mass of electron: 000000000000000000000000000910956 gramsRequires about 12 bytes

Floating Point NumbersDefinitions

Range How small and how large the numbers can be.

Precision The number of significant figures used to represent the

number. A measure of a number’s exactness.

PI = 3.141592 is more precise that PI = 3.14

Accuracy A measure of the correctness of a number.

PI = 3.241592 is more precise than PI = 3.14, but PI = 3.14 is more accurate.

IEEE Floating Point NumbersSingle Precision Format

-1s * 2E-B * 1.F

B = 127

IEEE Floating Point NumbersRange of Mantissa

A floating point mantissa is limited to one of the three ranges:

-2 < x <= -1x = 0+1 <= x < +2

IEEE Floating Point NumbersExponent

Binary Value True Exponent

Biased Exponent

Special Numbers

0000 0000 -127 0 zero

0000 0001 -126 1

0000 0010 -125 2

0000 0100 -124 3

. . .

1000 0000 0 128

. . .

1111 1100 125 252

1111 1101 126 253

1111 1110 127 254

1111 1111 128 255 +- Infinity

IEEE Floating Point NumbersExcess - n

The stored exponent is also called excess – n, or excess 127, for the IEEE single precision format.

The stored exponent exceeds the true exponent by 127, the bias.

b’ = b + 127where b’ is the biased exponent, and b is the true

exponent. Examples:

If the true exponent is 2, the exponent is stored in biased form as 2 + 127 = 1000 0001.

If the stored exponent is 0000 0001, the true exponent is 1 – 127 = -126.

IEEE Floating Point NumbersRepresentation of Zero

The smallest stored exponent 0000 0000 (in biased form), corresponding to a true exponent of -127, is used to represent zero.

IEEE Floating Point NumbersInfinity and Not a Number (NaN)

1111 1111 used as +- infinity.

1111 1111 and Mantissa != 0 used as NaN.

IEEE Floating Point NumbersExample Representation

Represent -2345.125 as a single precision IEEE floating point number.

-2345.12510 = -100100101001.0012

-2345.12510 = -1.001001010010012 x 211

S = 1 (negative) The biased exponent is 11 + 127 = 138 =

100010102 The fractional part of the mantissa

is .00100101001001000000000 Therefore, -2345.12510 =

1 10001010 00100101001001000000000

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IEEE Floating Point Numbers Arithmetic Example #1

1. Convert the decimal numbers 123.5 and 100.25 into the IEEE 32-bit floating point number representation. Then carry out the subtraction of 123.5 – 100.25 and express the result as a normalized 32-bit floating point number.

123.510 = 1111011.12 = 1.1110111 x 26

The mantissa is positive, and so S = 0. The exponent is +6, which is stored in biased

form as 6 + 127 = 13310 = 100001012. The mantissa is 1.1110111, which is stored in 23-

bits, with the leading ‘1’ suppressed. Therefore, 123.510 is stored as:

0 10000101 11101110000000000000000IEEE

IEEE Floating Point Numbers Arithmetic Example #1 (Continued)

1. Convert the decimal numbers 123.5 and 100.25 into the IEEE 32-bit floating point number representation. Then carry out the subtraction of 123.5 – 100.25 and express the result as a normalized 32-bit floating point number. (Continued)

100.2510 = 1100100.012 = 1.10010001 x 26

The mantissa is positive, and so S = 0. The exponent is +6, which is stored in biased

form as 6 + 127 = 13310 = 100001012. The mantissa is 1.10010001, which is stored in

23-bits, with the leading ‘1’ suppressed. Therefore, 100.2510 is stored as:

0 10000101 10010001000000000000000IEEE

IEEE Floating Point Numbers Arithmetic Example #1 (Continued)

1. Convert the decimal numbers 123.5 and 100.25 into the IEEE 32-bit floating point number representation. Then carry out the subtraction of 123.5 – 100.25 and express the result as a normalized 32-bit floating point number. (Continued)

The two IEEE numbers are first unpacked: the sign, exponent, and mantissa must be reconstituted.

The two exponents are compared. If they are the same, the mantissas are added. If they are not, the number with the smaller exponent is denormalized by shifting its mantissa right (i.e., dividing by 2) and incrementing its exponent (i.e., multiplying by 2) until the two exponents are equal. Then the numbers are added.

IEEE Floating Point Numbers Arithmetic Example #1 (Continued)

1. Convert the decimal numbers 123.5 and 100.25 into the 32-bit floating point number representation. Then carry out the subtraction of 123.5 – 100.25 and express the result as a normalized 32-bit floating point number. (Continued)

After unpacking, insert the leading ‘1’ and perform the subtraction.

1.11101110000000000000000 -1.10010001000000000000000 0.01011101000000000000000

Normalize the result: 1.01110100000000000000000

IEEE Floating Point Numbers Arithmetic Example #1 (Continued)

1. Convert the decimal numbers 123.5 and 100.25 into the IEEE 32-bit floating point number representation. Then carry out the subtraction of 123.5 – 100.25 and express the result as a normalized 32-bit floating point number. (Continued)

The exponent must be decreased by 2. 10000101 – 210 = 10000011

The result expressed in IEEE format is: 0 10000011 01110100000000000000000

IEEE Floating Point Numbers Arithmetic Example #2

2. Convert the decimal numbers 42.6875 and -0.09375 into the IEEE 32-bit floating point number representation. Then carry out the addition of 42.6875 and – 0.09375 and express the result as a normalized 32-bit floating point number.

42.687510 = 101010.10112 = 1.010101011 x 25

The mantissa is positive, and so S = 0. The exponent is +5, which is stored in biased

form as 5 + 127 = 13210 = 100001002. The mantissa is 1.010101011, which is stored in

23-bits, with the leading ‘1’ suppressed. Therefore, 42.687510 is stored as:

0 10000100 01010101100000000000000IEEE

IEEE Floating Point Numbers Arithmetic Example #2 (Continued)

2. Convert the decimal numbers 42.6875 and -0.09375 into the IEEE 32-bit floating point number representation. Then carry out the addition of 42.6875 – 0.09375 and express the result as a normalized 32-bit floating point number (continued).

-0.0937510 = -0.000112 = -1.1 x 2-4

The mantissa is negative, and so S = 1. The exponent is -4, which is stored in biased

form as -4 + 127 = 12310 = 011110112. The mantissa is 1.1, which is stored in 23-bits,

with the leading ‘1’ suppressed. Therefore, -0.0937510 is stored as:

1 01111011 10000000000000000000000IEEE

IEEE Floating Point Numbers Arithmetic Example #2 (Continued)

2. …+42.687510:0 10000100 101010101100000000000000

-0.0937510:1 01111011 110000000000000000000000

In order to perform the addition, the exponents must be the same.

Increase the second exponent by 9 and shift the mantissa right 9 times to get:

+42.687510:0 10000100 101010101100000000000000

-0.0937510:1 10000100 000000000110000000000000000000000

IEEE Floating Point Numbers Arithmetic Example #2 (Continued)

2. …+42.687510:0 10000100 101010101100000000000000

-0.0937510:1 10000100 000000000110000000000000000000000

Adding the mantissas, we get: 101010100110000000000000

The result is positive with a biased exponent of 10000100.

Therefore, the result is stored as:

0 10000100 0101010011000000000000