Decimal Addition

1
Assembly Language Programming CS208
2
Assembly Language

• Assembly language allows us to use convenient
  abbreviations (called mnemonics) for machine
  language operations and memory locations.
• Each assembly language is specific to a
  particular hardware architecture, and can only be
  used on a machine of that architecture.
• An assembly language program must be translated
  into machine code before it can be executed. The
  program that tells the computer how to perform
  the translation is called an assembler.

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Assembly Language

• When a processor chip is designed, it is designed
  to understand and execute a set of machine code
  instructions (OpCodes) unique to that chip.
• One step up from machine code is assembly code.
  Each machine code instruction is given a mnemonic
  (name), so that it is easier for human beings to
  write code.
• There is generally a one-to-one correspondence
  between the assembly languages mnemonic
  instructions and the machine language numeric
  instructions.

4
Model Assembly Instructions

• Our assembly language instructions have two
  parts
• The operation code specifies the operation the
  computer is to carry out (add, compare, etc)
• An address that allows the instruction to refer
  to a location in main memory
• The CPU runs each instruction in the program,
  starting with instruction 0, using the
  fetch-decode-execute cycle.

5
Review of the Fetch-Decode-Execute Cycle

• The CPU fetches the next instruction from the
  address contained in the Program Counter and
  places the instruction in the Instruction
  Register.
• When a program starts, the program counter
  contains 0, so the instruction at address 0 is
  fetched.
• Immediately after the instruction fetch, the CPU
  adds 1 word to the contents of the Program
  Counter, so that it will contain the address of
  the next sequential instruction.

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Review of theFetch-Decode-Execute Cycle

• The CPU decodes the instruction in the
  Instruction Register and determines what
  operations need to be done and what the address
  is of any operand that will be used.
• The specified operation is executed (add,
  compare, etc).
• After execution of the instruction has been
  completed the cycle starts all over again (unless
  the instruction terminates the program).

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CPU

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CPU Registers

• The Instruction Register (IR) contains the actual
  instruction which is currently being executed by
  the CPU.
• The Status Register records the result of
  comparing the contents of register A with the
  contents of register B.
• The Program Counter (PC) contains the address of
  the next instruction to be executed by the
  program.
•  

9
CPU Registers

• Registers A B hold the operands for each
  arithmetic operation (ie. the values on which the
  operation will be performed). After the operation
  has been carried out, the result is always stored
  in Register B.
• Therefore, after an arithmetic operation has been
  performed, the second operand is no longer stored
  in Register B, because it has been overwritten by
  the result of the operation.

10
CPU Registers

• After a comparison has been done, the Status
  Register will hold a code that stores the results
  of the comparison.
• The results are coded as follows
• -1 if (A lt B)
• 0 if (A B)
• 1 if (A gt B)

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Model Assembly Language Instructions

• Operation What it means to the CPU
• STP Stop the program
• LDA Load register A with value from
  a specified memory location
• LDB Load register B with value from
  a specified memory location
• STR Store register B value to a specified
  memory location
• INP Store data input by user to a specified
  memory location
• PNT Print the value stored in a specified
  memory location to the screen

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Model Assembly Language Instructions

• Operation What it means to the CPU
• JLT Jump if less than (Status register -1)
  to a specified memory location
• JGT Jump if greater than (Status register
  1) to a specified memory location
• JEQ Jump if equal (Status register 0) to a
  specified memory location
• JMP Unconditional jump to a specified memory
  location
• CMP Compare register A to register B and set
  Status Register value

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Model Assembly Language Instructions

• Operation What it means to the CPU
• ADD Add (register A register B) and store
  sum in register B
• SUB Subtract (register A - register B)
  and store difference in register B
• MUL Multiply (register A register B) and
  store product in register B
• DIV Divide for quotient (register A/register
  B) and store quotient in register B
• MOD Divide for remainder (register A/register
  B) and store remainder in register B

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Steps to write Assembly Programs

• Create C Program (only the statements between
  the and brackets are needed)
• Translate each C statement to the equivalent
  assembly statement(s)
• Number the assembly language program starting
  from 0
• Replace memory names by memory address numbers of
  empty memory cell
• Resolve jumps (replace with number of memory cell
  jumping to)

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C to Assembly Language

• Statement Assembly equivalent
• include none
• void main() none
• const value in memory cell
• int, double, char address of memory cell
• cin INP
• cout PNT
• assignment () LDA Val1 val3 val1
  val2 LDB Val2 ADD STR Val3
• STP

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Sample Program 1

• Program 1.
• Write an assembly language program that will get
  a number as input from the user, and output its
  square to the user.

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Sample Program 1

• Step 1
• Write an algorithm to describe the steps needed
  to solve our problem.
• Algorithm
• 1. Input a number and store it in memory.
• 2. Compute the square by multiplying the
  number times itself.
• 3. Output the results.

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Sample Program 1

• Step 2 Write the C code
• int Number , Square
• cout ltlt "Enter a number "
• cin gtgt Number
• Square Number Number
• cout ltlt Square

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Sample Program 1 Step 3 Translate C code
to assembly

• cout ltlt "Enter a number "
• cin gtgt Number
• square number number
• cout ltlt Square

• INP number
• LDA number
• LDB number
• MUL
• STR square
• PNT square
• STP

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Sample Program 1

• Step 4 Number assembly code lines starting
  from 0
• 0 INP number
• 1 LDA number
• 2 LDB number
• 3 MUL
• 4 STR square
• 5 PNT square
• 6 STP

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Sample Program 1

• Step 5 Replace memory names with empty memory
  locations after STP
• 0 INP number 7
• 1 LDA number 7
• 2 LDB number 7
• 3 MUL
• 4 STR square 8
• 5 PNT square 8
• 6 STP

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Sample Program 1

• Step 6 Final Assembly code
• INP 7
• LDA 7
• LDB 7
• MUL
• STR 8
• PNT 8
• STP

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Running the Code on the Model Assembler

• Type the code on the previous slide into a file
  (use Notepad).
• Save the file as sample1.txt in the same
  directory as the assembler.exe file
• Double click assembler.exe
• Press ENTER
• Type the filename sample1.txt
• Press r to run

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Before running the code, the screen will look
like this
Your Assembly Code
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After running the code, the screen will look like
this
Results of Program Run
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C Decisionsto Assembly Language

• C Statement
• if ( Num lt 10 )
• cout ltlt Num
• Assembly equivalent
• LDA Num
• LDB Ten
• CMP test condition
• JLT Then block address
• JMP address of statement after Then block
• PNT Num Then block

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C Decisions to Assembly Language

• Pascal Statement
• if ( Num lt 10 )
• cout ltlt Num
• else
• cout ltlt 0
• Assembly equivalent
• LDA Num
• LDB Ten
• CMP Test condition
• JLT Then block address
• PNT Zero Else block
• JMP Address of statement after Then block
• PNT Num Then block

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Sample Program 2

• Program 2.
• Write an assembly program that will get a number
  from the user, and determine if the number is
  evenly divisible by 5.
• Output zero (false) if the number is NOT evenly
  divisible by 5 or one (true) if the number IS
  evenly divisible.

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Sample Program 2

• Step 1 Write the algorithm to describe the steps
  needed to solve our problem.
• 1. Read in a number and store it in memory.
• 2. Determine if input number is evenly divisible
  by 5.
• 2.1 Divide input number by 5 to get the
  remainder.
• 2.2 Compare remainder to 0.
• If remainder equals 0,
• the number is evenly divisible.
• If the remainder does not equal 0,
• the number NOT evenly divisible.
• 3. Output the results
• 3.1 If evenly divisible, output 1.
• 3.2 If NOT evenly divisible, output 0.

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Sample Program 2

• Step 2 Write the C code
• const int Zero 0
• const int One 1
• const int Five 5
• int number, rem
• cout ltlt "Enter number "
• cin gtgt number
• rem number Five
• if (rem Zero)
• cout ltlt One
• else
• cout ltlt Zero

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Sample Program 2

• Step 3 Translate C code to Assembly
• cout ltlt "Enter number "
• cin gtgt number INP number
• LDA number
• rem number Five LDB Five
• MOD
• STR rem

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Sample Program 2

• Step 3 continued
• if (rem Zero) LDA Zero
• cout ltlt One LDB rem
• else CMP
• cout ltlt Zero JEQ then block address
• PNT Zero ?else block
• JMP address after then block
• PNT One ?then block
• STP

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Sample Program 2

• Step 4 Number assembly code lines starting
  from 0
• 0 INP number
• 1 LDA number
• 2 LDB Five
• 3 MOD
• 4 STR rem
• 5 LDA Zero ?condition
• 6 LDB rem
• 7 CMP
• 8 JEQ then block address
• 9 PNT Zero ?else block
• 10 JMP address after then block
• 11 PNT One ?then block
• 12 STP

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Sample Program 2

• Step 5 Replace names by cell numbers after STP
• 0 INP number 16
• 1 LDA number 16
• 2 LDB Five 13
• 3 MOD
• 4 STR rem 17
• 5 LDA Zero 14
• 6 LDB rem 17
• 7 CMP
• 8 JEQ then block address
• 9 PNT Zero 14
• 10 JMP address after then block
• 11 PNT One 15
• 12 STP
• 13 5 ? Five
• 14 0 ? Zero
• 15 1 ? One
• 16 ? number
• 17 ? rem

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Sample Program 2

• Step 5 Replace jumps by instruction numbers
• 0 INP 16
• 1 LDA 16
• 2 LDB 13
• 3 MOD
• 4 STR 17
• 5 LDA 14
• 6 LDB 17
• 7 CMP
• 8 JEQ address of then block 11
• 9 PNT 14 ?else block
• 10 JMP address after then block 12
• 11 PNT 15 ?then block
• 12 STP
• 13 5
• 14 0
• 15 1
• 16
• 17

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Sample Program 2

• Step 6 Final Assembly code
• INP 16
• LDA 16
• LDB 13
• MOD
• STR 17
• LDA 14
• LDB 17
• CMP
• JEQ 11
• PNT 14
• JMP 12
• PNT 15
• STP
• 5
• 0
• 1

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C Decisions to Assembly Language

• C Statement
• while ( Num lt 10 )
• cout ltlt Num
• Assembly equivalent
• LDA Num
• LDB Ten
• CMP ? test condition
• JLT to While Block
• JMP to stmt after While Block
• PNT Num ? While Block
• JMP to test condition
• ? stmt after While Block

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Sample Program 3

• Program 3.
• Write a program to display a count by fives to
  100.

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Sample Program 3

• Step 1 Write an algorithm to describe the steps
  needed to solve our problem
• 1. Set Count to start at 0
• 2. While Count is less than 100
• 2.1 Add 5 to the Count and store the sum back
  into the Count
• 2.2 Display the Count

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Sample Program 3

• Step 2 Write C code
• const int Five 5
• const int Hundred 100
• int Count 0
• while (Count lt Hundred)
• Count Count 5
• cout gtgt Count

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Sample Program 3

• Step 3 Translate C code to assembly
• while (Count lt Hundred)
• Count Count 5
• cout gtgt Count
• LDA Count ? test condition
• LDB Hundred
• CMP
• JLT to while block
• JMP to stmt after while block
• LDB Five ? while block
• ADD
• STR Count
• PNT Count
• JMP to test condition
• ? stmt after while block

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Sample Program 3

• Step 4 Number assembly code lines from 0
• 0 LDA Count ? test condition
• 1 LDB Hundred
• 2 CMP
• 3 JLT to while block
• 4 JMP to stmt after while block
• 5 LDB Five ? while block
• 6 ADD
• 7 STR Count
• 8 PNT Count
• 9 JMP to test condition
• 10 STP ? stmt after while block

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Sample Program 3
Step 5 Replace memory names with empty memory
locations after STP 0 LDA Count 13 1 LDB
Hundred 12 2 CMP 3 JLT to while block 4 JMP to
stmt after while block 5 LDB Five
11 6 ADD 7 STR Count 13 8 PNT Count
13 9 JMP to test condition 10 STP 11 5 ?
Five 12 100 ? Hundred 13 0 ? Count
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Sample Program 3
Step 5 Replace memory names with empty memory
locations after STP 0 LDA 13 ? test
condition 1 LDB 12 2 CMP 3 JLT to while block
5 4 JMP stmt after while block 10 5 LDB 11 ?
while block 6 ADD 7 STR 13 8 PNT 13 9 JMP to
test condition 0 10 STP ? stmt after while
block 11 5 12 100 13 0
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Sample Program 3
Final Assembly Code LDA 13 LDB 12 CMP JLT
5 JMP 10 LDB 11 ADD STR 13 PNT 13 JMP
0 STP 5 100 0