Assembly Language Programming
CPU
CPU The CPU contains a Control Unit, Arithmetic Logic
Unit (ALU) and a small number of memory locations called Registers.
Different registers perform different tasks such:
– as manipulating data, – keeping track of the results of decision making operations, – and pointing to the next instruction to be executed.
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.
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.
CPU Registers The computer also has a Compare instruction that
can be used to compare the contents of register A with those of register B.
The comparison can have three possible outcomes: – the contents of register A < B; – the contents of the register A = B; – the contents of the register A > B.
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 < B);• 0 if (A = B); • 1 if (A > B).
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.
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 a one-to-one correspondence between the assembly languages mnemonic instructions and the machine language numeric instructions.
A list of assembly code instructions that will perform a task is called an assembly program.
Assembly language.Operation What it means to the CPU
STP Stop the program
LDA Load register A with contents of a specified memory location
LDB Load register B with contents of a specified memory location
STR Store register B contents to a specified memory location
INP Store data input by user to a specified memory location
PNT Print the contents of a specified memory location to the screen
Assembly language.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
Assembly language.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) andstore 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
Steps to write Assembly Programs Create Pascal Program translate each Pascal statement to the equivalent
assembly statement(s) Number the assembly language program starting
from 0 Replace Memory names by number of empty
memory cell Resolve jumps ( replace with number of memory
cell jumping to)
Pascal to Assembly language.Statement Assembly equivalent
program none
const put value in memory cell
var put address of memory cell
readln INP
writeln PNT
assignment (:=)
val3 = val1 + val2 LDA Val1
LDB Val2
ADD
STR Val3
End. STP
Assembly language.Program #1.
Write an assembly language program that will get a number as input from the user, and output its square to the user.
Assembly language.step 1: algorithm to describe the steps needed to solve
our problem.
Assembly language.step 1: algorithm to describe the steps needed to solve
our problem.
1. Input a number and store it in memory.
2. Compute the square by multiplying
the number times itself.
3. Output the results.
Assembly language.Step 2: write Pascal code
Assembly language.Step 2: write Pascal code
Var
number , square: integer;
begin
readln ( number);
square := number * number ;
writeln (square);
end.
Assembly Languagebegin
readln ( number);
square := number*number;
writeln (square);
end.
INP number
LDA number
LDB number
MUL
STR square
PNT square
STP
Assembly language.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
Assembly language.Step 5: Replace memory names by cell numbers 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
Assembly language.Step 6: Final Assembly code
INP 07
LDA 07
LDB 07
MUL
STR 08
PNT 08
STP
Pascal to Assembly language.Statement Assembly equivalent
if ( N < 10 ) then LDA N (condition expression)
writeln (N) LDB Ten
CMP
JOP (operation) (Then Block)
JMP to statement after then block
PNT N ( then block)
Pascal to Assembly language.Statement Assembly equivalent
if ( N < 10 ) then LDA N (condition expression)
writeln (N) LDB Ten
else writeln (‘0’); CMP
JOP (operation) (Then Block)
PNT Zero ( else Block)
JMP to statement after then block
PNT N ( then block)
Assembly language.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.
Assembly language.step 1: algorithm to describe the steps needed to solve
our problem.
Assembly language.step 1: algorithm to describe the steps needed to solve our
problem.
1. Input a number and store it in memory.
2. Determine if the input number is evenly divisible by 5.
2.1Divide the input number by 5 to get the remainder.
2.2 Compare the 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.
Assembly language.Step 2: write Pascal code
Assembly language.Step 2: write Pascal code
Const
Zero = 0; One =1; Five = 5;
Var
number , rem: integer;
begin
readln ( number);
rem := number MOD Five ;
if (rem = Zero) then
writeln (One);
else writeln (Zero)
end.
Assembly language.Step 3: translate Pascal code to assembly
Const
Zero = 0; One =1; Five = 5;
Var
number , rem: integer;
begin
readln ( number); INP number
rem := number MOD Five ; LDA number
LDB Five
MOD
STR rem
Assembly language.Step 3: translate Pascal code to assembly
if (rem = Zero) then condition exp LDA Zero
writeln (One); LDB rem
else writeln (Zero) CMP
end. JEQ then block
else block PNT Zero
JMP after then
then block PNT One
STP
Assembly language.Step 4: Number assembly code lines starting from 0
0 INP number1 LDA number2 LDB Five3 MOD4 STR rem
condition exp 5 LDA Zero6 LDB rem7 CMP8 JEQ then block
else block9 PNT Zero10 JMP after then
then block 11 PNT One12 STP
Assembly language.Step 5: Replace memory names by cell numbers after STP
0 INP number 161 LDA number 162 LDB Five 133 MOD4 STR rem 17
condition exp 5 LDA Zero 146 LDB rem 177 CMP8 JEQ then block
else block9 PNT Zero 1410 JMP after then
then block 11 PNT One 1512 STP
13 514 015 1
Assembly language.Step 5: Replace jumps by instruction numbers
0 INP 161 LDA 162 LDB 133 MOD4 STR 175 LDA 146 LDB 177 CMP8 JEQ then block 11
else block9 PNT 1410 JMP after then 12
then block 11 PNT 1512 STP13 514 015 1
Assembly language.Step 6: Final Assembly code
INP 16LDA 16LDB 13MODSTR 17LDA 14LDB 17CMPJEQ 11
PNT 14JMP 12PNT 15STP501
Assembly language.Statement Assembly equivalent
While (N < 10 ) do LDA N (condition expression)
begin LDB Ten
writeln (N) CMP
end; JOP (operation) (While Block)
JMP to statement after while block
PNT N ( while block statements)
JMP Condition
Assembly language. Program #3. Write an assembly program that will add up a series of
positive numbers entered by the user, until the user enters a negative number, then display the total.
Assembly language.step 1: algorithm to describe the steps needed to solve
our problem.
Assembly language.step 1: algorithm to describe the steps needed to solve
our problem.
1. Input a value and store it in memory. 2. While the Input Value is not a negative
number:– 2.1 Add the Input Value to the Running Total and store – the sum back into the Running Total.– 2.2 Input another value and store it in memory.
3. Output the contents of the Running Total.
Assembly language.Step 2: write Pascal code
Assembly language.Step 2: write Pascal code
Const
Zero = 0; Var
sum , number: integer;begin
sum := zero;readln ( number);While ( number >= Zero) dobegin
sum := sum + number;readln (number);
end; writeln (sum)
end.
Assembly language.Step 3: translate Pascal code to assembly
begin
sum := zero;
readln ( number);
While ( number >= Zero) do
LDB Zero
STR sum
INP number
condition LDA numberLDB Zero
CMP
JGT (While Block)
JEQ (While Block)
JMP after while block
Assembly language.Step 3: translate Pascal code to assembly
begin
sum := sum + number;
readln (number);
end;
writeln (sum)
end.
LDA sumLDB numberADDSTR sum
INP number
JMP Condition
PNT sum
STP
Assembly language.Step 4: Number assembly code lines starting from 0
0 LDB Zero
1 STR sum
2 INP number
condition 3 LDA number4 LDB Zero
5 CMP
6 JGT (While Block)
7 JEQ (While Block)
8 JMP after while block
9 LDA sum10 LDB number11 ADD12 STR sum
13 INP number
14 JMP Condition
15 PNT sum
16 STP
Assembly language.Step 5: Replace memory names by cell numbers after
STP0 LDB Zero 17
1 STR sum 18
2 INP number 19
condition 3 LDA number 194 LDB Zero 17
5 CMP
6 JGT (While Block)
7 JEQ (While Block)
8 JMP after while block
while body 9 LDA sum 1810 LDB number 1911 ADD12 STR sum 18
13 INP number 19
end while 14 JMP Condition
15 PNT sum 18
16 STP17 0
Assembly language.Step 5: Replace jumps by cell numbers
0 LDB 17
1 STR 18
2 INP 19
condition 3 LDA 19
4 LDB 17
5 CMP
6 JGT (While Block) 9
7 JEQ (While Block) 9
8 JMP after while block 15
while body 9 LDA 1810 LDB 1911 ADD12 STR 18
13 INP 19
end while 14 JMP Condition 3
15 PNT 18
16 STP17 0
Assembly language.Step 6: Final Assembly code
LDB 17
STR 18
INP 19
LDA 19
LDB 17
CMP
JGT 9
JEQ 9
JMP 15
LDA 18LDB 19ADDSTR 18INP 19JMP 3PNT 18STP0
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