monday, october 4 assignment(s) due: assignment #5: complement arithmetic assignment #6: debug...

Monday, October 4

Assignment(s) due: Assignment #5: COMPLEMENT ARITHMETICAssignment #6: DEBUG EXERCISES

Quiz(zes) due: Quiz #5: Base ArithmeticQuiz #6 (Complement Arithmetic)

Note: Quiz #7 (Debug) will be due next Monday

Emulator Progress Worksheet 1 will be due next Monday

Today is the cutoff date for all Simple Computer labs and assignments

And next Monday is the cutoff date for all arithmetic assignments and quizzes(Assignment #5, Quiz #5, Quiz #6, Optional Quizzes #5, #7 and #8)

Please note: The date of the midterm is Monday, October 18- an 8.5 in. * 11 in. cheat sheet will be allowed

We will review for the midterm next week

I recommend going over the midterm practice exercises before I discuss them on Monday

Don't forget the optional quiz: Debug

Given the following information in DEBUG:

-e 120 34 12 00 01

-u 100, 10A21F0:0100 BB2001 MOV BX, 12021F0:0103 BE0200 MOV SI, 221F0:0106 0307 MOV AX, [BX]21F0:0108 0300 ADD AX, [BX + SI]

Trace through the partial program above, showing the contents of the AX, BX and SI registers

AX BX SI

-e 120 34 12 00 01

AX BX SI

But first:

What does this mean? 120 121 122 123

-e 120 34 12 00 01

AX BX SI

But first:

What does this mean? 120 121 122 123 34 12 00 01

-e 120 34 12 00 01

AX BX SI ? 0120 ?

-e 120 34 12 00 01

AX BX SI ? 0120 ? ? 0120 0002

-e 120 34 12 00 01

AX BX SI ? 0120 ? ? 0120 00021234 0120 0002

-e 120 34 12 00 01

AX BX SI ? 0120 ? ? 0120 00021234 0120 00021334 0120 0002

Review:

MOV CX ,0003MOV AX, 0000ADD AX, CX

LOOP 106

INT 20

Using the LOOP instruction (works with CX):

How does the loop instruction work?

Review:

MOV CX ,0003MOV AX, 0000ADD AX, CX

LOOP 106

INT 20

Using the LOOP instruction (works with CX):

How does the loop instruction work?

1. It decrements CX

2. It checks CX (is CX > 0?)

3. If CX > 0- it loops to the address specified

If CX = 0- it goes to the next instruction in sequence

- each location in the computer is assigned a unique address

- memory locations- input/output devices

- the address allows them to be recognized by the CPU

- the CPU puts the address on the address bus - identified by the decoding circuits

- the data bus is used to get data to or from the address

- the control bus sends read/write signals- to indicate if the CPU is asking for or sending information

- every process in the computer (at every level of the computer)- uses this fetch/decode/execute cycle

- each location in the computer is assigned a unique address

- memory locations

- input/output devices

- the address allows them to be recognized by the CPU

- the CPU puts the address on the address bus

- identified by the decoding circuits

- the data bus is used to get data to or from the address

- the control bus sends read/write signals

- to indicate if the CPU is asking for or sending information

(ALU,registers,controlunit)

RAM ROM Printer cd/dvd Monitor Keyboard Mouse

Address Bus

Data Bus

Control Bus

- the function (or purpose) of the CPU:- to fetch and execute instructions

- more efficient if the process is divided

- at the same time:- one instruction is being executed, and- the next instruction is being fetched

- implemented by theoretically dividing the CPU into:

- the execution unit- decodes and executes instructions

and- the bus interface unit

- fetches the instructions

- the execution unit contains- the ALU

- performs operations- arithmetic- logical

- registers- high speed memory- temporarily store data or addresses

- general purpose registers- available to programmers

- special purpose registers- e.g. the instruction pointer, the instruction register- not available to programmers- to indicate if the CPU is asking for or sending information

- the flags register- is set (or reset) as the result of operations done in the ALU- can be checked by a program

- the decoder- determines the type of instruction- interprets it into a form that can be used in the ALU and registers

- the bus interface unit- where instructions and data come in from main memory

- the bus interface unit delivers them to the execution unit

- contains:- the instruction queue

- allows the bus interface unit to stay ahead of the execution unit- when the queue is full

- the buses are idle- if the program performs a jump

- the queue will be changed

- the instruction pointer- holds the address of the next instruction to be executed

- the address will be placed on the address bus

- the segment registers- hold the addresses of the various parts of memory used by the program

- where the instructions are stored- where the data is stored- where the program’s stack is located

- the execution unit and the bus interface unit work in parallel- but the bus interface unit is always at least one step ahead

Registers:

General

- EAX (accumulator)

EAX - 32 bits

8 hexadecimal digits

AX - 16 Bits

4 hexadecimal digits

AH and AL - 8 bits each

2 hexadecimal digits each

- EBX, ECX, EDX

- and BX, BH, BL, CX, CH, CL, DX, DH, DL

Intel Registers (grouped by function)

Data Registers(general purpose)

Index Registers(used in string operations)

Pointer Registers(used in stack operations)

Accumulator register

Base register

Count register

Data register

Source Index register

Destination Index register

Stack Pointer register

Base Pointer register

Segment Registers (16-bit)(hold segment addresses)

Control RegistersFlags register detail:

- bit 0 CF- bit 2 PF- bit 6 ZF- bit 7 SF- bit 10 DF- bit 11 OF

Intel Registers (grouped by function)

Code Segment register

Data Segment register

Stack Segment register

Extra Segment register

EFLAGS

Instruction pointer register

Flags register

- immediate addressing

- the operand comes immediately after the instruction

e.g. MOV EAX, 12345678 or MOV AX, ABCD or MOV AH, 2A

- direct addressing

- specifies the address of the operand

e.g. MOV EAX, [0100] or MOV EAX, NUMBER

- register addressing

- specifies the register in which the operand is stored

e.g. MOV EAX, EBX

32-bit Addressing Modes

Scientific Notation Format

NormalizedStandard Form Scientific Notation Scientific Notation

3.14159 3.14159 x 100 .314159 x 101

100,000 1.0 x 105 .1 x 106

.0001 1.0 x 10-4 .1 x 10-3

-1234 -1.234 x 103 -.1234 x 104

-1.234 3 -.1234 4

mantissa characteristic fraction exponent (or significand)

Scientific notation makes it relatively easy to work with very large or very small numbers- normalized scientific notation used in the computer is called floating point notation

4-Decimal-Digit Floating Point Format

Fraction2 decimal digitsNormalizedSigned

Exponent2 decimal digitsSigned

3.14159

100,000

( .314159 * 101)

( .1 * 106)

( .1 * 10-3)

( .1234 * 104)

4-Decimal-Digit Floating Point Format

3.14159

100,000

Range: - defined by the number of digits in the exponent (-.99 x 10+99 .. +.99 x 10+99)

Precision: - defined by the number of digits in the fraction (2 digits of precision)

- if the 4-decimal digit format is changed:

Exponent1 decimal digitSigned

- the range is smaller (-.999*109 .. 999*109)

- there is a smaller distance between adjacent numbers- there is greater accuracy

Fraction1 decimal digitNormalizedSigned

- the range is increased

- accuracy is greatly decreased

Fraction1 decimal digitNormalizedSigned

- the range is increased

- accuracy is greatly decreased

- no matter what format is used- these numbers will generate 7 different regions of storage:

- a positive value region, - a negative value region and a - zero region- 2 overflow regions - 2 underflow regions

Range and regions of value for a 4-decimal-digit floating point format:

zero .10x100

.10x10-99 .99x1099-.10x10-99-.99x1099

-.10x100

Range of 4-decimal-digit floating point format: (using the number line)

(Absolutely not to scale)

overflow overflow

underflow

Roundoff Error

If we store 3.14159 as we truncate (round down) to 3.1

If we store 3.14159 as we round up to 3.2

If we store 1234 as we round down to 120041 2 0

Floating point formats:

- often stored in normalized scientific notation form(.xxxxxx * base exponent)

- precision (accuracy)- affected by the number of digits in the fraction- affected by the limits of binary representation

- range- determined by the number of digits in the exponent

- often generate a round-off error

Binary Floating Point Formats

- most are based on dividing the binary number into 3 parts: - the sign (1 bit)

- 0 = positive, 1 = negative- the exponent

- uses excess-N notation- the fractional part

- will be normalized- the radix point

- will be to the left or right of the most significant bit- depending on the format used

A 32-bit floating point number:

To calculate the fraction:

- we must know the position of the radix point - to the left or right of the most significant bit

- we must know the position of the radix point- to the left or right of the most significant bit (it's to the left)

- and the base that will be used to calculate the exponent- called the radix of exponentiation

- usually base 2, 8 or 16

- usually base 2, 8 or 16 (we'll look at all three)

To calculate the exponent:

- we must first normalize the fraction to find the exponent- using the radix of exponentiation

To calculate the exponent:

- we must first normalize the fraction to find the exponent- using the radix of exponentiation

- if the exponent is stored in 7 bits (it's stored in 7 bits)- it uses excess-64 notation

- if the exponent is stored in 8 bits- it uses excess-127 notation

The problem:

- how 0.00110 would be represented in a hypothetical machine- using a 32-bit floating point format

- 1 bit for the sign bit - (0 = positive, 1 = negative)

- 7 bits for the exponent - using excess-64 notation

- 24 bits for the fraction- normalized

- radix point to the left of the most significant bit

- the radix of exponentiation is 2

Step 1: Convert 0.00110 to radix 2

0.00110 = 0.00040611156458

= 0.0000000001000001100010010011011101001012

0.00110 = 0.00040611156458

= 0.0000000001000001100010010011011101001012

Step 2: Normalize the fraction

- move the radix point to the left of the most significant bit

- i.e. move the radix point to the right 9 bits

= 0.0000000001000001100010010011011101001012

0.00110 = 0.00040611156458

= 0.0000000001000001100010010011011101001012

- move the radix point to the right 9 bits

= 0.0000000001000001100010010011011101001012

- on a computer using radix 2 the normalized form is:

= 0.100000110001001001101110 1001012 * (2-9)10

0.00110 = 0.00040611156458

= 0.0000000001000001100010010011011101001012

= 0.100000110001001001101110 1001012 * (2-9)10

the 24 bits we need for the fraction

- normalizing the fraction gives us the fraction in radix 2 and the exponent in radix 10

= 0.100000110001001001101110 1001012 * (2-9)10

fraction exponent

= 0.100000110001001001101110 1001012 * (2-9)10

fraction exponent

Step 3: Convert the exponent to its excess-64 form in radix 2

- in radix 10: -9 + 64 = 5510

- in radix 2: 01101112

= 0.100000110001001001101110 1001012 * (2-9)10

fraction exponent

- in radix 10: -9 + 64 = 5510

- in radix 2: 01101112

Step 4: Put the number together

sign exponent fraction

0 0110111 100000110001001001101110or

3 7 8 3 1 2 6 E in hexadecimal

The problem:

- the radix of exponentiation is 8- for radix 8 we move the radix point in sets of 3 bits (1 octal digit = 3 binary digits)

0.00110 = 0.0000000001000001100010010011011101001012

- move the radix point to the right in sets of 3 bits

- and stay to the left of the msb

= 0.0000000001000001100010010011011101001012

0.00110 = 0.0000000001000001100010010011011101001012

= 0.0000000001000001100010010011011101001012

= 0.100000110001001001101110 1001012 * (8-3)10

Note: the roundoff error is truncation error

- if the roundoff error was from rounding bit #31 would be a 1

= 0.100000110001001001101110 1001012 * (8-3)10

fraction exponent

- in radix 10: -3 + 64 = 6110

- in radix 2: 01111012

= 0.100000110001001001101110 1001012 * (8-3)10

fraction exponent

- in radix 10: -3 + 64 = 6110

- in radix 2: 01111012

0 0111101 100000110001001001101110or

3 D 8 3 1 2 6 E in hexadecimal

The problem:

- the radix of exponentiation is 16- for radix 16 we move the radix point in sets of 4 bits (1 hexadecimal digit = 4 binary

digits)

(This was the short-form floating point format used on the IBM 370 computer)

0.00110 = 0.0000000001000001100010010011011101001012

= 0.0000000001000001100010010011011101001012

0.00110 = 0.0000000001000001100010010011011101001012

= 0.0000000001000001100010010011011101001012

= 0.010000011000100100110111 01001012 * (16-2)10

Note: the roundoff error is is the same for rounding or truncation

- the next bit is a 0

= 0.010000011000100100110111 01001012 * (16-2)10

fraction exponent

- in radix 10: -2 + 64 = 6210

- in radix 2: 01111102

= 0.010000011000100100110111 01001012 * (16-2)10

fraction exponent

- in radix 10: -2 + 64 = 6210

- in radix 2: 01111102

0 0111110 010000011000100100110111or

? in hexadecimal

Conversion to/from floating point formats:

- decimal number to hexadecimal floating point format:

- convert decimal number to binary

- convert to normalized scientific form (varies)

- convert exponent to its excess-N form in binary

- add sign bit to exponent and fraction

- convert binary number to hexadecimal

To decode the hexadecimal number 3E400000 (same format):

The number in binary: 0 011 1110 0100 0000 0000 0000 0000 0000

0 6210 .012

Sign: positive

Exponent: -210

Fraction: .012

0 6210 .012

Sign: positive

Exponent: -210

Fraction: .012

The normalized form: .01 x (16-2) 10

0 6210 .012

Sign: positive

Exponent: -210

Fraction: .012

The non-scientific binary form: .00000000012

0 6210 .012

Sign: positive

Exponent: -210

Fraction: .012

The non-scientific binary form: .00000000012

The number in decimal: = 1/1024 or .0009765...10

- hexadecimal floating point format to decimal value

- convert hexadecimal number to binary

- determine sign, exponent and fraction

- convert exponent to its decimal form

- convert to normalized form

- convert to its non-scientific binary form

- convert to decimal

The problem from Lab #9:

- how -46.7410 would be represented in a hypothetical machine- using a 32-bit floating point format

digits)

The problem from Lab #9:

digits)

.74 * 2 = 1.48

.48 * 2 = 0.96

.96 * 2 = 1.92

.92 * 2 = 1.84

.84 * 2 = 1.68

.68 * 2 = 1.36

.36 * 2 = 0.72

.72 * 2 = 1.44

.44 * 2 = 0.88

.88 * 2 = 1.76

.76 * 2 = 1.52

.52 * 2 = 1.04

.04 * 2 = 0.08

.08 * 2 = 0.16

.16 * 2 = 0.32

.32 * 2 = 0.64

.64 * 2 = 1.28

.28 * 2 = 0.56

.56 * 2 = 1.12

Step 1: Convert -46.7410 to radix 2

-46.7410 = -101110.10111101011100001012

Step 2: Normalize the fraction- move the radix point to the left of the most significant bit

- move the radix point to the left in sets of 4 bits- and stay to the left of the msb

= -00101110.10111101011100001012

= -.0010111010111101011100002 * (162)10

Step 3: - the roundoff error is is from rounding - the next bit is a 1

- the normalized form becomes

= -.0010111010111101011100012 * (162)10

-46.7410 = -101110.1011110101110000101012

= -00101110.1011110101110000102

= -.0010111010111101011100002 * (162)10

= -.0010111010111101011100012 * (162)10

-46.7410 = -101110.101111010111000010102

= -00101110.1011110101110000102

= -.0010111010111101011100002 * (162)10

= -.0010111010111101011100012 * (162)10

fraction exponent

- in radix 10: 2 + 64 = 6610

- in radix 2: 10000102

1 1000010 .001011101011110101110001or C 2 2 E B D 7 1 in hexadecimal

= -.0010111010111101011100012 * (162)10

fraction exponent

- in radix 10: 2 + 64 = 6610

- in radix 2: 10000102

= -.0010111010111101011100012 * (162)10

fraction exponent

- in radix 10: 2 + 64 = 6610

- in radix 2: 10000102

IEEE Standard Floating Point Formats for the Intel 8087 microprocessor:

- the Short Real form: (uses excess-127 for the exponent)

IEEE Standard Floating Point Formats for the Intel 8087 microprocessor:

- the Short Real form: (uses excess-127 for the exponent)

- the Long Real form: (uses excess-1023 for the exponent)

The IEEE Short Real form: (and the Long Real form)

- divides the binary number into 3 parts:

- the sign (1 bit)- 0 = positive, 1 = negative

- the exponent-uses excess-N notation

- excess-127 for Short Real- excess-1023 for Long Real

-the radix of exponentiation is 2

- the exponent-uses excess-N notation

- excess-127 for Short Real- excess-1023 for Long Real

-the radix of exponentiation is 2

- the fractional part- normalized using the scientific notation method

- with the radix point to the right of the most significant bit (always a 1)- then the msb is implied in the number and not stored

- it's considered "hidden" in the floating point format

More IEEE Standard Floating Point Formats for the Intel 8087 microprocessor

- the Temporary Real form: (uses excess-32767 for the exponent)

- the temporary real form: - uses normalized notation

- msb to the right of the binary point and NOT hidden

- there are also 4 additional formats to handle overflow and underflow

The problem:

- how 0.00110 would be represented in a hypothetical machine- using a 32-bit floating point format (IEEE standard)

- radix point to the right of the most significant bit- msb is "hidden"

- the radix of exponentiation is 2

0.00110 = 0.00040611156458

= 0.0000000001000001100010010011011101001012

- move the radix point to the right of the most significant bit

= 0.0000000001000001100010010011011101001012

0.00110 = 0.00040611156458

= 0.0000000001000001100010010011011101001012

- move the radix point to the right of the most significant bit

= 0.0000000001000001100010010011011101001012

= 1.00000110001001001101110 1001012 * (2-10)10

fraction exponent

- in radix 10: -10 + 127 = 11710

- in radix 2: 011101012

= 1.00000110001001001101110 1001012 * (2-10)10

fraction exponent

- in radix 10: -10 + 127 = 11710

- in radix 2: 011101012

0 01110101 00000110001001001101110or

3 A 8 3 1 2 6 E in hexadecimal

To decode the hexadecimal number 3D400000 (IEEE format):

The number in binary: 0 011 1110 0 100 0000 0000 0000 0000 0000

0 12410 .12

Sign: positive

Exponent: -3

Fraction: 1.1

0 12410 .12

Sign: positive

Exponent: -3

Fraction: 1.1

The normalized form: 1.1 x (2-3)10

0 12410 .12

Sign: positive

Exponent: -3

Fraction: 1.1

The binary number: .00112

0 12410 .12

Sign: positive

Exponent: -3

Fraction: 1.1

The binary number: .00112

The decimal number: 1/8 +1/16 or 3/16 or .187510

- hexadecimal floating point format to decimal value

- convert hexadecimal number to binary

- determine sign, exponent and fraction

- convert exponent to its decimal form

- convert to normalized form

- convert to its non-scientific binary form

- convert to decimal

Monday, October 4

Assignment(s) due: Assignment #5: COMPLEMENT ARITHMETICAssignment #6: DEBUG EXERCISES

Quiz(zes) due: Quiz #5: Base ArithmeticQuiz #6 (Complement Arithmetic)

Note: Quiz #7 (Debug) will be due next Monday

Emulator Progress Worksheet 1 will be due next Monday

Today is the cutoff date for all Simple Computer labs and assignments

And next Monday is the cutoff date for all arithmetic assignments and quizzes(Assignment #5, Quiz #5, Quiz #6, Optional Quizzes #5, #7 and #8)

Please note: The date of the midterm is Monday, October 18- an 8.5 in. * 11 in. cheat sheet will be allowed

We will review for the midterm next week

I recommend going over the midterm practice exercises before I discuss them on Monday

Don't forget the optional quiz: Debug

The problem from Lab #9 (Part 2):

- radix point to the right of the most significant bit- and the most significant bit is "hidden"

- the radix of exponentiation is 2- for radix 2 we move the radix point bit by single bit

(This is the IEEE standard short-form floating point format)

.74 * 2 = 1.48

.48 * 2 = 0.96

.96 * 2 = 1.92

.92 * 2 = 1.84

.84 * 2 = 1.68

.68 * 2 = 1.36

.36 * 2 = 0.72

.72 * 2 = 1.44

.44 * 2 = 0.88

.88 * 2 = 1.76

.76 * 2 = 1.52

.52 * 2 = 1.04

.04 * 2 = 0.08

.08 * 2 = 0.16

.16 * 2 = 0.32

.32 * 2 = 0.64

.64 * 2 = 1.28

.28 * 2 = 0.56

.56 * 2 = 1.12

-46.7410 = -101110.10111101011100001012

Step 2: Normalize the fraction- move the radix point to the right of the most significant bit

- move the radix point to the left bit by bit

= -00101110.10111101011100001012

= -1.011101011110101110000102 * (25)10

= -1.011101011110101110000112 * (25)10

-46.7410 = -101110.10111101011100001012

= -00101110.10111101011100001012

= -1.011101011110101110000102 * (25)10

= -1.011101011110101110000112 * (25)10

-46.7410 = -101110.10111101011100001012

= -00101110.10111101011100001012

= -1.011101011110101110000102 * (25)10

= -1.011101011110101110000112 * (25)10

fraction exponent

- in radix 10: 5 + 127 = 13210

- in radix 2: 100001002

1 10000100 .01110101111010111000011or C 2 3 A F 5 C 3 in hexadecimal

= -1.011101011110101110000112 * (25)10

fraction exponent

- in radix 10: 5 + 127 = 13210

- in radix 2: 100001002

= -1.010111010111101011100012 * (25)10

fraction exponent

- in radix 10: 5 + 127 = 13210

- in radix 2: 100001002

monday, october 4 assignment(s) due: assignment #5: complement arithmetic assignment #6: debug...

bx 21f0

bx si trace

debug slide

si registers axbxsi

partial program

following information

mov ax

debug exercises quizzes

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