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Write an Assembly program that reads in a number of cents.

Customer Question

Write an Assembly program that reads in a number of cents. The program will write out the number of dollars and cents separately.

Sample Run:
? 324
3
24

Write an Assembly program to read three numbers in from the user, and output BOTH the sum and the product of the three numbers.

Sample Run:
? 2
? 3
? 4
9
24

Write an Assembly program to determine the cost for a group to attend a movie. Tickets are $6 for children and $8 for adults. Read in the number of children and the number of adults, and display the total movie cost.

Sample Run:
? 2
? 1
20

Write an Assembly program to compute a person's gas mileage. Prompt for the beginning mileage, ending mileage, and the number of gallons (to the nearest tenth) it took to fill the car. Compute the gas mileage (whole miles per gallon).

Sample Run:
? 32000
? 32250
? 10
25
Submitted: 4 years ago.
Category: Programming
Expert:  DRouleau replied 4 years ago.
hello, I can assist you but I need more information.

What assembly language? For a specific processor of generic? Can you provide a sample?

When do you need it by?

Regards,

Denis
Customer: replied 4 years ago.
i think C+, it is intro to Comp Sci
Expert:  DRouleau replied 4 years ago.
C++ is not an assembly language.

Do you have a sample for you text book? If I don't have an example I won't be able to write it in the right format for you and you will have to translate it. Not really desirable.

Regards

Denis
Customer: replied 4 years ago.
Invitation to Computer Science - C++ Version, 4th Edition. Thompson Course Technology
Expert:  DRouleau replied 4 years ago.
Thanks for the book title. Unfortunately, I don't have access to this book to look at examples. Can you send me a picture or scan of a pages that shows an example?

Otherwise, I'm going to have to pass.

Regards,

Denis
Customer: replied 4 years ago.

This is sample pages

A computer performs various tasks by executing a sequence of

instructions that the processor is able to understand. 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 is a language that allows total control over

every individual machine instruction generated by the assembler.

High-level language compilers, on the other hand, make many

decisions about how a given language statement will be translated

into machine instructions. For example, the following single C++

instruction assigns a value of 55 to a numeric variable called Num:

Num = 55;

When the C++ compiler reads this line, it outputs a series of four

or five machine instructions that take the value 55 and store it in

memory at a location named Num. Normally, you the programmer

have no idea what these four or five instructions actually are, and

you have no way of changing them, even if you know a sequence

of machine instructions that is faster and more efficient than the

sequence the compiler uses. Assembly language gives you the

choice of which machine instructions will be used.

Assembly Language Concepts

  • Each individual instruction is very simple. A single instruction

rarely does more than move a word of data from one storage area

to another, compute a value, or compare the value contained in one

storage area to a value contained in another. So you can

concentrate on the simple task that is being accomplished by the

instruction without worrying about the complexity of the entire

program.

  • A program requires many instructions to do anything useful. You

can often write a useful program in a high-level language (like

C++) in five or six lines. But a functional assembly language

program cannot be implemented in fewer than about twenty lines,

and anything demanding takes hundreds or thousands of lines.

  • The key to assembly language is understanding memory

addresses. In high-level languages like C++, the compiler takes

care of where the data is located-you simply have to give the

constant or variable data a name, and call it by that name when you

want to use it. But in assembly language, you must always be

aware of where things are in your computer's memory. All data

will be accessed via memory addresses.

Our Model Assembly Language

We will use an assembly language code that consists of 16

instructions. The instruction set is listed below.

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

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

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

Notice that instructions LDA, LDB, STR, INP, PNT, JLT, JGT,

JEQ, and JMP all require a "specified memory address". The

complete instruction actually consists of two parts: an opcode and

an operand. The opcode is one of the operations listed above and

the operation is a memory address.

The load and store instructions. The load instructions load

values (i.e. copy the contents) from main memory locations into

registers A and B. There is one load instruction for register A,

LDA, and one for register B, LDB. The store instruction, STR,

stores a value (i.e. copies the contents) from the register B into

main memory location.

Examples

The instruction: LDA 20

loads the contents of memory location 20 into the register A.

The contents of memory location 20 are not affected by the

copying; its value remains the same.

The instruction: LDB 21

loads the contents of memory location 21 into the register B.

The contents of memory location 21 are not affected by the

copying; its value remains the same.

The instruction: STR 22

stores the contents of register B into memory location 22. The

contents of register B are not affected by the copying; its value

remains the same.

The input and output instructions. The input instruction, INP,

transfers data entered by the user at the computer's single input

device (the keyboard) to a main memory location. The output

instruction, PNT, displays the data contents of a main memory

location on the computer's single output device (the monitor

screen).

Examples

The instruction: INP 23

reads one number entered at the keyboard and stores it in

memory location 23.

The instruction: PRT 23

displays the number in memory location 23. The contents of

location 23 remain unchanged.

The arithmetic instructions. The arithmetic instructions are

ADD, SUB, MUL, DIV, and MOD. There are two division

instructions, DIV and MOD. DIV yields the integer quotient of a

division, while MOD yields the remainder. For each instruction,

one of the numbers on which the operation is to be carried out

must be in register A; the other must be in register B. The result

of the operation is always stored back into register B. Since the

operands must be stored in the registers, these operations do not

require a memory address.

For addition and multiplication, the contents of the registers A and

B are added/multiplied (i.e. register A + register B).

For subtraction, the contents of register B are subtracted from the

contents of register A (i.e. register A - register B).

For division, the contents of register B are divided into the contents

of register A (i.e. register A / register B).

Examples

The instruction: ADD

adds contents registers A and B. The sum is stored in the

register B.

The instruction: SUB

subtracts the contents of register B from the contents of

register A. The difference is stored in register B.

The instruction: MUL

multiplies the contents of registers A and B. The product is

stored in register B.

The instruction: DIV

divides the contents of register A by the contents of register

B. The whole quotient is stored in register B.

The instruction: MOD

divides the contents of register A by the contents of register

B. The remainder of the division is stored in register B

The compare instruction. The compare instruction and the jump

instructions work together to implement selection and repetition.

The compare instruction, CMP, compares the contents of the

register A with the contents of register B.

The instruction: CMP

compares the contents of register A with the contents of

register B. It records in the Status Register whether register

A's contents are less than, equal to, or greater than Register

B's contents, according to the following chart:

-1 if the contents of register A are less than those

of register B (A < B);

0 if the contents of registers A and B are the same

(A = B);

1 if the contents of register A are greater than those

of register B (A > B).

The jump instructions. The unconditional jump instruction,

JMP, changes the value of the program counter, by storing a new

memory address into it. This causes the computer to take its next

instruction from the newly specified address. The computer jumps

to the point in the program specified in the jump instruction and

continues execution from that point.

The instruction: JMP 15

causes the computer to jump to memory location 15 and

continue program execution from there. In the absence of

further jump instructions, the computer will execute the

instructions in locations 16, 17, 18, and so on.

JLT, JEQ, and JGT are conditional jump instructions. The jump

only takes place if the contents of the Status Register have a

particular value. If the contents of the Status Register do not have

the required value, the jump does not take place and the computer

continues execution of the program with the instruction in memory

following the one that holds the jump instruction.

JLT jumps only if the contents of the status register are -1

(indicating register A was less than register B)

JEQ jumps only if the contents are 0 (A was equal to B)

JGT jumps only if the contents are 1 (A was greater than B).

Examples

The instruction: JLT 11

causes the computer to jump to location 11 in the program

only if the contents of the status register are -1 (register A's

contents were less than register B's contents when last

compared). If the comparison code register contains 0 or 1,

the computer continues with the instruction in the memory

location following the one containing the jump instruction

(no jump).

The instruction: JGT 12

causes the computer to jump to location 12 in the program

only if the contents of the status register are 1 (A was greater

than B).

The instruction: JEQ 11

causes the computer to jump to location 11 in the program

only if the contents of the status register are 0 (A was equal

to B).

If the condition for one of these jumps is true, then the address

after the jump instruction is loaded into the program counter,

which causes the jump to a new instruction.

The stop instruction. The stop instruction, STP, causes the CPU to stop

fetching and executing instructions and go into standby mode. The

address part of the instruction is ignored.

The instruction: STP

brings the program execution to a stop.

Type the operand (i.e. memory address) for instructions

that require one in columns 5-6.

To create program data values:

Data values will be placed on the lines immediately

following the STP operation in your program.

Beginning in column 1 of each line, type the value you

wish to store in that memory location.

The Model Assembler

You will be using a model computer assembler. See section 5.2.2

in your week-by-week for instructions on loading the assembler to

your PC.

Running Programs

After starting the assembler (double click on the .exe file), you can

run assembly programs. Enter the name of the file containing the

source code at the filename prompt. The program will be loaded

into memory, and displayed on the screen. The program counter

contains 0, the address of the first instruction, and all other

registers are empty.

You can now run the program in either NORMAL or STEP mode.

NORMAL mode - Press 'r' to run (execute) the entire program.

The program will run, starting with the instruction at address

0, and will continue to run until termination, pausing only to

get input from the user.

STEP mode - Press any key to execute ONE step at a time.

The program begins to run, starting with the instruction at address

0. Each line of your program code will take three steps to

complete. Each time you press a key, one step will execute. The

steps are as follows:

1) Fetch the instruction stored at the RAM address specified in

the program counter and place it in the instruction register.

2) Increment the program counter by 1 (to point to the next

instruction address).

3) Decode and execute the instruction.

The program display will show the user how the computer loads

and performs the instructions within the program, as well as

showing a simulated monitor to display the results of the program

input/output.

SAMPLE PROGRAMS

Sample Program #1.

Our first sample program will get a number as input from the user,

and output its square to the user. First we need an English-like

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. Store the result in memory.

4. Output the results from memory.

Since our sample program will use two memory locations to store

data, as follows:

What's Stored Memory Address

Input Number 07

Squared Result 08

Step 1 of the algorithm inputs the number from the user and stores

it in location 07.

INP 07

Step 2 computes the square of the number at location 07. First the

number at location 07 is loaded into both registers. Then the

register values are multiplied together. The result ends up in

register B.

LDA 07

LDB 07

MUL

Step 3 stores the result (currently in register B) into memory

location 08.

STR 08

Step 4 displays the contents of memory location 08 to the screen

and then terminates the program.

PNT 08

STP

Putting all the instructions together, the program is shown below.

Memory

Address Contents Explanation

00 INP 07 Get input number from user and store at location 07

01 LDA 07 Load input number into register A

02 LDB 07 Load input number into register B

03 MUL Multiply register A by register B, result in B

04 STR 08 Store result from register B into location 08

05 PNT 08 Output result stored in location 08

06 STP Terminate execution

07 ? Storage for Input Number

08 ? Storage for Squared Result

Sample Program #2.

Our second sample program will get a number from the user, and

determine if the number is XXXXX divisible by 5. Output zero

(false) if the number is XXXXX evenly divisible by 5 or one (true) if

the number IS XXXXX divisible.

Again, we first we need an English-like algorithm to describe the

steps needed to solve our problem.

1. Input a number and store it.

2. Determine if the input number is XXXXX divisible by 5.

2.1 Divide the input number by 5 to get the remainder.

2.2 Compare the remainder to 0.

2.2.1 If remainder equals 0, the number IS XXXXX divisible.

2.2.2 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

his sample program will use four memory locations to store data.

What's Stored Address

Zero 11

One 12

Five 13

Input number 14

Step 1 of the algorithm inputs the number from the user and stores

it in location 14.

INP 14

Step 2 determines if the number is XXXXX divisible by 5. This is

accomplished by loading the register A with the input number

(from location 14), loading register B with value 5 (from location

13), and MODing the registers. Then the results of the MOD are

compared with zero, by loading 0 into register A (from location

11) and comparing it to the remainder of the MOD in register B.

LDA 14

LDB 13

MOD

LDA 11

CMP

Step 3 takes the comparison results and outputs the answer. First,

if the comparison showed the MOD result in register B was equal

to 0 telling us the number is XXXXX divisible by 5, then the

program jumps to output 1 (from location 12). Otherwise, the

number is XXXXX evenly divisible, and the program outputs zero

(from location 11).

JEQ 09

PNT 11

If 0 was output, the program must then jump to the stop statement,

to prevent outputting of 1 also.

JMP 10

PNT 12

STP

Putting all the instructions together, the program is shown below.

Memory

Address Contents Explanation

00 INP 14 Input number and store in location 14

01 LDA 14 Load value at location 14 into register A (input)

02 LDB 13 Load value at location 13 into register B (5)

03 MOD MOD register A value (input) by register B value (5)

04 LDA 11 Load value at location 11 into register A (0)

05 CMP Compare register A (0) to register B (mod result)

06 JEQ 09 If condition code 0 (values equal), jump to instruction 09

07 PNT 11 Output value at location 11 number (input number)

08 JMP 10 Jump to instruction 10

09 PNT 12 Output value at location 12 (0)

10 STP Terminate execution

11 0 Storage for constant 0

12 1 Storage for constant 1

13 5 Storage for constant 5

14 ? Storage for input number

Sample Program #3.

Our third sample program will add up a series of positive numbers

entered by the user, until the user enters a zero, and then displays

the total.

Below is the English-like algorithm to solve the problem.

1. Copy the value 0 into the Running Total.

2. Input a value and store it in memory.

3. While the Input Value is not zero

3.1 Add the Input Value to the Running Total and store the

sum back into the Running Total.

3.2 Input another value and store it in memory.

4. Output the contents of the Running Total.

The program will use three memory locations to store data:

What's Stored Address

Zero 13

User input value 14

Running Total 15

Step 1 of the algorithm clears the Running Total by copying 0 into

the Running Total. This is done by loading 0 (from location 13)

into the register B and storing it into the Running Total (at

location 15).

LDB 13

STR 15

Step 2 gets the first input value to be processed and stores it in

location 14

INP 14

Steps 3.1 and 3.2 are to be repeated only so long as the Input Value

is NOT equal to zero. Before executing these steps, the computer

must compare the contents of the Input Value with 0. If the

contents of the Input Value is 0, the computer is to jump around

steps 3.1 and 3.2 and go directly to step 4. If the Input Value is

NOT equal to 0, the computer is to continue with steps 3.1 and 3.2.

The program loads the Input Value (from location 14) into the

register A, and 0 (from location 13) into register B, then compares

the contents of the two registers.

LDA 14

LDB 13

CMP

If the Input Value in register A is equivalent to the 0 in register B,

the computer should jump to step 4, the instructions for which

begin at memory location 11.

JEQ 11

If the Input Value in register A was NOT equal to the 0 in register

B, the JEQ instruction has no effect, and the computer goes on to

the instructions for steps 3.1 and 3.2. Step 3.1 adds the Input

Value (at location 14) to the Running Total (at location 15) and

stores the result back into the Running Total. The Input Value has

already been loaded into the register A; thus we must load the

Total into register B and use an ADD instruction to add the

contents of the two registers.

LDB 15

ADD

The STR instruction stores the resulting sum in register B back

into the Running Total (at location 15).

STR 15

Steps 3.1 and 3.2 are to be repeated while the contents of the Input

Value are NOT equal to zero. This means that when step 3.2 has

been executed, the computer must jump back to the beginning of

step 3. Step 3 will check the new Input Value, jumping to step 4 if

the Input Value is zero, or continuing with steps 3.1 and 3.2

otherwise. When steps 3.1 and 3.2 have been executed, the

computer will again jump back to the beginning of step 3, where

the next Input Value will be checked, and so on. An unconditional

jump instruction causes the computer to jump back to the

beginning of step 3.

JMP 02

Step 4 outputs the Running Total (at location 15) and terminates

the program.

PNT 15

STP

Putting all the instructions together, the program is shown below.

Address Contents Explanation

00 LDB 13 Load 0 into Register B (from location 13)

01 STR 15 Store 0 into Total (initialize location 15)

02 INP 14 Read user Input Value (into location 14)

03 LDA 14 Load Input Value (from location 14) into register A

04 LDB 13 Load 0 (from location 13) into register B

05 CMP Compare contents of register A (input value) and register B (0)

06 JEQ 11 Jump out of loop if contents of A equal contents of B (0)

07 LDB 15 Load the Total into register B

08 ADD Add register A contents (input value) to register B contents (total)

09 STR 15 Store resulting sum (in register B) to location 15 (total)

10 JMP 02 Jump back to start of loop for next input

11 PNT 15 Output total (from location 15)

12 STP Terminate program execution

13 0 Storage for constant 0

14 ? Storage for Input Value

15 ? Storage for Total

5.2.1.4 C++ VERSUS ASSEMBLY LANGUAGE

Now that we have some Assembly fundamentals AND some C++

fundamentals, let's look at how the summation program from the

previous section (sample program #3) could be handled by C++.

NOTES FOR PROGRAM.

1. The repetition of the cout and cin statements before and in the

"while" loop allow the loop to terminate immediately if the first value

entered is zero (0) since the condition is tested BEFORE the loop is

"executed". If the loop is entered, the value is added to total and the user

is again asked to enter a value. This data entry and summation is

repeated until a zero (0) is entered from the keyboard.

2. Note the use of "{" and "}" to mark the boundaries of the "while"

loop. All statements between these curly braces are controlled by the

loop. In other words, they are executed only as long as the loop is active.

Once the condition for looping becomes FALSE, these statements will

no longer execute.

5.2.2 Using the Model Assembler

Double-click the ASSEMBLER.EXE file

The following screen should appear:

Customer: replied 4 years ago.
Expert:  DRouleau replied 4 years ago.
Perfect, I can work with that.

When do you need it?

Regards,

Denis
Customer: replied 4 years ago.
12-03 at 5pm mst
Expert:  DRouleau replied 4 years ago.
DONE. They are all in this file.

I couldn`t test them but I`m sure you`ll be able to.

Regards

Denis
Customer: replied 4 years ago.

i'll test

Expert:  DRouleau replied 4 years ago.
Don't forget to click on Accept afterwards. That's how I get paid for my service.

Best regards,

Denis
Customer: replied 4 years ago.
http://academic.regis.edu/mlotfy/cs208/computer_fundamentals_online.htm#model%20assembler
Customer: replied 4 years ago.

Denis

Thanks a lot, this is the link to download the assemble,can you check http://academic.regis.edu/mlotfy/cs208/computer_fundamentals_online.htm#model%20assembler

Expert:  DRouleau replied 4 years ago.
Hi,
I tested them and all is good. Use these final versions.

prog1.zip
prog2.zip
prog3.zip
prog4.zip

Regards XXXXX XXXXX't forget to accept;)

Denis
DRouleau, Computer Software Engineer
Category: Programming
Satisfied Customers: 106
Experience: C * Python * Assembler * Javascript * PHP * SQL * Linux * Shell * Basic * Java *
DRouleau and 5 other Programming Specialists are ready to help you
Customer: replied 4 years ago.
Great, I have 4 more similar programs, would you have time, they are due 12-5 6pm mst
Customer: replied 4 years ago.

These are actually programs3 not 4

 

Write an Assembly program to determine the correct admission price to a movie. The price of admission to a movie is $7 for kids (under 12) and $9 for adults. Display the correct admission fee after the user gives you the age for one person. Sample Run #1:
? 21
9
Sample Run #2:
? 11
7

 

 

Write an Assembly program to determine the price of a car rental. The car being rented costs $45 per day and frequent renters get a $15 discount on the total bill. Get as input, the number of days rented and whether the user is part of the frequent renter program (have user enter 1 if they ARE a frequent renter and 0 otherwise). Display the total cost to rent the car.

Sample Run #1:
? 4
? 0
180 Sample Run #2:
? 3
? 1
120

 

Write an Assembly program to output a list of number. Read from the user a starting value and an ending value. Your program should output all integers from the starting value through the ending value.

Sample Run:
? 5
? 8
5
6
7
8

Expert:  DRouleau replied 4 years ago.
I can do them now but they will have to be part of an other question.

Title the question with my name in it. For example: For drouleau: 3 more programs.

Regards,

Denis
Customer: replied 4 years ago.

drouleau: 3 more programs

These are actually programs3 not 4

 

Write an Assembly program to determine the correct admission price to a movie. The price of admission to a movie is $7 for kids (under 12) and $9 for adults. Display the correct admission fee after the user gives you the age for one person. Sample Run #1:
? 21
9
Sample Run #2:
? 11
7

 

 

Write an Assembly program to determine the price of a car rental. The car being rented costs $45 per day and frequent renters get a $15 discount on the total bill. Get as input, the number of days rented and whether the user is part of the frequent renter program (have user enter 1 if they ARE a frequent renter and 0 otherwise). Display the total cost to rent the car.

Sample Run #1:
? 4
? 0
180 Sample Run #2:
? 3
? 1
120

 

Write an Assembly program to output a list of number. Read from the user a starting value and an ending value. Your program should output all integers from the starting value through the ending value.

Sample Run:
? 5
? 8
5
6
7
8

Expert:  DRouleau replied 4 years ago.
HiCustomer
what I meant is that you have to open a new question as this is new work not covered by the original question.

Please open a new question and title it as I previously described.

Best regards,

Denis

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