07 - number representations

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EEM336

Microprocessors I

Representation of Numbers

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Contents

Representation of Numbers

Unsigned, Sign magnitude, Onescomplement, Twos complement, Biased

Real Numbers

Single Precision, Double Precision

BCD Numbers

ASCII Numbers

Shift and Rotate

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Integer Representation

There are 5 commonly known integerrepresentations:

1. Unsigned

2. Sign magnitude

3. Ones complement

4. Twos complement

5. Biased (not commonly known)

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Integer Representation(continued)

All have been used at various timesfor various reasons.

Virtually all modern computersoperate based on twos complementrepresentation. Because,

Hardware is faster

Hardware is simpler (which makes itfaster)

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Unsigned Numbers

The standard binary encoding thatyou already know

Only positive integer numbers arerepresented

Range: 0 to 2n - 1 for n bits

Examples: 5 is represented as 0000 0101

134 is represented as 1000 0110

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Sign Magnitude

Both positive and negative integerscan be represented

The most significant bit (MSB) is used

to represent the sign For positive numbers, MSB is 0 and

for negative numbers, MSB is 1

Examples: 5 is represented in 8-bit as 0000 0101

-5 is represented in 8-bit as 1000 0101

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Ones Complement

Positive integers use the samerepresentation as unsigned

Negation is done by taking a bitwisecomplement of the positiverepresentation

Examples:

5 is represented in 8-bit as 0000 0101


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Twos Complement

Positive integers use the samerepresentation as unsigned

Negation is done by taking the onescomplements and adding 1

Examples:

5 is represented in 8-bit as 0000 0101


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Biased Representation

A bias value is chosen as either 2n-1 or 2n-1-1,according to the representation algorithm

The integer number to be represented is added tothe bias

Unsigned representation of the result is the biasedrepresentation of the number

Examples:

In 8-bit, if bias is 127, then 5 is 1000 0100 (132)

In 8-bit, if bias is 127, then -5 is 0111 1010 (122)

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Sign Extension in TwosComplement

Assume that you have a signed number in 8 bits andyou want to convert it to 16 bits

This type of conversion from smaller number of bits tolarger number of bits is called sign extension

In this case, the original number is copied into LSBsand the sign bit is copied into the MSBs

Examples:5: 0000 0101 in 8 bits 0000 0000 0000 0101 in 16 bits

-5: 1111 1010 in 8 bits 1111 1111 1111 1010 in 16 bits

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CharacterRepresentation

ASCII codes are used to represent the characters

Characters are generally represented in 8-bits

Examples:

A is represented as 0100 0001 (65 or 41h)

B is represented as 0100 0010 (66 or 42h)

0 is represented as 0011 0000 (48 or 30h)

9 is represented as 0011 1001 (57 or 39h)

In Assembly, characters should be written into single-quotes (i.e.dont use double-quotes)

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Converting from Stringto Integer

The 0 character and the number 0 aredifferent

If you want to convert the string 354 to

integer, you should follow this algorithm: Read 3 and translate it to 3 (just subtract

30h)

Read 5 and translate it to 5

integer = 3 * 10 + 5 = 35

Read 4 and translate it to 4

integer = 35 * 10 + 4 = 354

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Representation of RealNumbers

Real numbers (or floating-point numbers)are often encountered in computer programs

A real number should be converted to binarysystem and encoded into bits

There are several representations but Intelmicroprocessors use IEEE 754, version 10.0standard

Intel 8086 has no floating point instructions, so

you have to code them by yourself

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Converting Real Numbersinto Binary

Separate the number into integerand fractional parts

You can easily convert the integerpart into binary

You can convert the fractional partby multiplying it by 2 continuouslyand taking the integer parts of theresults

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Example: Convert 64.2 toBinary

Separate it into 64 and 0.2

Binary representation of 64 is 100 0000

0.2 can be gotten using the algorithm:

0.2 * 2 = 0.4 0 (MSB) 0.4 * 2 = 0.8 0

0.8 * 2 = 1.6 1

0.6 * 2 = 1.2 1

0.2 * 2 = 0.4 0 (and it repeats itself) 0.2 = 0.0011 0011 0011

Then, 64.2 is 100 0000.0011 0011 0011

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Scientific Notation

The real numbers should benormalized and written in scientificnotation

In scientific notation, a number hasthree components: Sign

Mantissa

Exponent

In normalized form, the integer part ofthe mantissa should be 1.

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Example: ScientificNotation of 64.2

64.2 can be written in binary as:

1.00 0000 0011 0011 0011 26

Here, 1.00 0000 0011 0011 0011 is called the mantissa

6 is the exponent

Sign is plus (+)

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IEEE 754 Standard

According to the IEEE 754 standard,floating-point numbers can berepresented in either 32 bits or 64

bits

32 bit version is called single-precision and 64 bit version is

called double-precision

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Single Precision

The first bit (bit 31) represents the sign (0 forpositive and 1 for negative numbers)

Next 8 bits (23-30) represent the biasedexponent(bias = 127)

Next 23 bits (0-22) represent the 24 bits of themantissa (24 bits in 23 bits!!! Is it correct?)

Yes! The first bit of the mantissa, which comes from theinteger part, is not needed to be coded

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S Exponent Mantissa

31 30 23 22 0


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Double Precision

The first bit (bit 63) represents the sign (0 forpositive and 1 for negative numbers)

Next 11 bits (52-62) represent the biasedexponent(bias = 1023)

Next 52 bits (0-51) represent the 53 bits of themantissa (53 bits in 52 bits!!! Is it correct?)

Yes! The first bit of the mantissa, which comes from theinteger part, is not needed to be coded

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S Exponent Mantissa

63 62 52 51 0


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64.2 in Single Precision

Since 64.2 is positive, sign bit should be 0

Exponent is 6; add 127 and obtain biased value (6+ 127 = 133 = 1000 0101)

Mantissa is 1.00 0000 0011 0011 0011

Ignore 1 in integer part and take the 23 bits fromthe fractional part: 000 0000 0110 0110 01100110

Combine them: 0100 0010 1000 0000 0110 01100110 0110 (or 42806666h)

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64.2 in Double Precision

Since 64.2 is positive, sign bit should be 0

Exponent is 6; add 1023 and obtain biased value (6 +1023 = 1029 = 100 0000 0101)

Mantissa is 1.00 0000 0011 0011 0011

Ignore 1 in integer part and take the 52 bits from thefractional part: 0000 0000 1100 1100 1100 1100 11001100 1100 1100 1100 1100 1100

Combine them: 0100 0000 0101 0000 0000 1100 11001100 1100 1100 1100 1100 1100 1100 1100 1100 (or40500CCCCCCCCCCCh)

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Exceptions

Zero is stored as all zeros.

The number infinity is stored as allones in the exponent and all zeros in

the mantissa. The sign bit indicateseither a positive or a negative infinity.

+ = 01111 1111 000 0000 0000

0000 0000 0000- = 11111 1111 000 0000 0000

0000 0000 0000

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Exercises

Calculate the single and double precisionrepresentations of12.3 (ans: 4144 CCCDh and4028 9999 9999 999Ah)

Calculate the single and double precisionrepresentations of-23.4 (ans: C1BB 3333hand C037 6666 6666 6666h)

What does the single precision floating point

number 3F80 0000h represent?

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BCD (Binary CodedDecimal) Data

In some cases, you might considerrepresenting each digits of an integernumber in 4 bits

For example, 1234h might representthe decimal value 1234

These type of numbers are called

BCD numbers There are some special instructions

to be used with BCD numbers

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DAA

DAA: Decimal Adjust after Addition

Corrects the result of addition of twopacked BCD values on AL register

Example:

MOV AL, 15 ; AL = 0Fh

DAA ; AL = 15h

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DAA Example

Sum two 4-digit BCD numbers stored inBX and DX and store the result in CX asa BCD.MOV DX, 1234h ; Load 1234 BCD

MOV BX, 3099h ; Load 3099 BCD MOV AL, BL ; Sum low bytes

ADD AL, DL ; AL = 34h + 99h = CDh

DAA ; CF = 1, AL = 33h (133)

MOV CL, AL ; Store low byte of result

MOV AL, BH ; AL = 30h (sum high bytes)

ADC AL, DH ; AL = 43h

DAA ; CF = 0, AL = 43h

MOV CH, AL ; Store high byte of result

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DAS

DAS: Decimal Adjust afterSubtraction

Corrects the result of subtraction oftwo packed BCD values on ALregister

Similar to DAA but DAS follows asubtraction

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ASCII Coded Numbers

You may use ASCII codes of numbersto represent them

For example, you can represent 19 as

3139h since ASCII codes of 1 and 9are 31h (=49) and 39h (=57)respectively

There are some instructions to be usedon ASCII coded numbers: AAA, AAD,AAM, and AAS.

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AAA

AAA: ASCII Adjust after Addition

Corrects result in AH and AL afteraddition of two ASCII coded numbers

Works on AL and changes AL and AH 3030h should be added to AX after AAA

Example:MOV AX, '01' ; AX = 3031h (ASCII code for 1)

ADD AL, '9' ; AL = 0Ah

AAA ; AH = 01h, AL = 00h

ADD AX, 3030h ; AX = 3130 (ASCII code for 10)

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Other ASCII CodeOperations

AAD: ASCII Adjust before Division

AAM: ASCII Adjust afterMultiplication

AAS: ASCII Adjust after Subtraction

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Shift and Rotate

Shift and rotate instructions manipulatebinary numbers at the binary bit level

Common applications in low-level

software used to control I/O devices Shift/Rotate count may be an immediate

value or CL

Examples:

SHL AX, 1SHR BX, CL

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Shift

Position or move numbers to the left orright within a register or memorylocation.

also perform simple arithmetic asmultiplication by powers of 2+n (left shift) anddivision by powers of 2-n (right shift).

The microprocessors instruction set

contains four different shift instructions: two are logical; two are arithmetic shifts

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The Shift Instructions

logical shifts move0 in the rightmostbit for a logical leftshift;

0 to the leftmost bitposition for a logicalright shift

arithmetic rightshift copies thesign-bit through the

number logical right shift

copies a 0 throughthe number.

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Logical vs. Arithmetic?

Logical shifts multiply or divideunsigneddata; arithmetic shiftsmultiply or divide signeddata.

a shift left always multiplies by 2 foreach bit position shifted

a shift right always divides by 2 for each

position shifting a two places, multiplies or

divides by 4

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Rotate

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References

http://pages.cs.wisc.edu/~smoler/x86te

emu8086 Emulators Documentation

Textbook

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http://pages.cs.wisc.edu/~smoler/x86text/TOC.lectnotes.htmlhttp://pages.cs.wisc.edu/~smoler/x86text/TOC.lectnotes.html

07 - number representations

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