Logical Instructions in 8085 Microprocessor (Solved Problem 2)

Summary

In this educational video, the presenter continues a series on logical instructions in 8085 assembly language programming. The session focuses on a problem where data from memory location X is multiplied by four and stored at location Y. The solution involves using the RLC instruction twice to achieve the multiplication. The process is demonstrated with a practical example, including writing the program, setting memory locations, and executing it in an emulator. The result is a clear explanation of how to manipulate data in assembly language.

Takeaways

• 📘 The session focuses on solving problems based on logical group of instructions in 8085 assembly language.
• 🔢 The specific problem involves taking data from memory location X, multiplying it by four, and storing the result at memory location Y.
• 🕊 The RLC (Rotate Left Circular) instruction is used for multiplying the accumulator content by two, which can be executed twice to achieve a multiplication by four.
• 📚 The ldax instruction is utilized to load data from memory location X into the accumulator.
• 🔄 Executing the RLC instruction twice effectively shifts the bits in the accumulator left by two positions, doubling the value twice (2x2=4).
• 📝 After the multiplication, the sty instruction is used to store the result back into memory location Y.
• 🏁 The HLT (Halt) instruction signifies the end of the program, signaling to the microprocessor that the program has concluded.
• 💾 The example uses memory locations 440 for input data (X) and 4410 for output data (Y), with the input data being '02' in hexadecimal.
• 🔧 The emulator is used to assemble, load, and execute the program, demonstrating the process of taking input, performing operations, and producing output.
• 📈 The input data '02' when multiplied by four results in '08', which is the expected output stored in memory location 4410 after program execution.
• 🎓 The session concludes with a recap of the covered topic and a teaser for the next session, which will involve more logical instructions.

Q & A

What is the main topic of the session?

The main topic of the session is to solve a problem on logical group of instructions using 8085 assembly language.

What is the specific problem that the session aims to solve?

The session aims to solve a problem where data from memory location X is taken, multiplied by four, and then stored at memory location Y.

What is the role of the RLC instruction in the 8085 assembly language in this context?

The RLC (Rotate Left Circular) instruction is used to multiply the data in the accumulator by two. Executing it twice will multiply the data by four.

How does the RLC instruction work in the context of this problem?

The RLC instruction rotates the content of the accumulator to the left by one bit. Executing it twice effectively multiplies the data by 2 x 2, which is 4.

What data transfer instruction is used to load data from memory location X into the accumulator?

The data transfer instruction 'ldax' is used to load data from memory location X into the accumulator.

What data transfer instruction is used to store the result in memory location Y?

The 'sty' instruction is used to store the result in memory location Y.

What is the purpose of the HLT instruction in the program?

The HLT (Halt) instruction is used to signal the microprocessor that the program has ended.

What is the input data stored at memory location 4400 in the example provided?

The input data stored at memory location 4400 is 02 in hexadecimal, which is 6 in decimal.

What is the expected output data after the program execution in the example?

The expected output data after the program execution is 08 in hexadecimal, which is 8 in decimal, as the input data is multiplied by 4.

How does the bit sequence change after executing the RLC instruction twice on the input data 02?

After executing the RLC instruction twice on the input data 02, the bit sequence changes from 0010 to 1000, which is the binary representation of 8.

What is the significance of the emulator in this session?

The emulator is used to assemble, load, and run the 8085 assembly language program to verify the correctness of the solution.

Outlines

00:00

🤖 Assembly Language Programming: Multiplying Data by Four

This paragraph introduces a session focused on solving problems using logical group instructions in 8085 assembly language. The specific task is to write a program that retrieves data from memory location X, multiplies it by four, and stores the result in memory location Y. The explanation begins by recalling the RLC instruction, which multiplies the accumulator's content by two, and then demonstrates how executing it twice achieves the required multiplication by four. The process involves loading the data from X into the accumulator using 'ldax', applying 'RLC' twice, and storing the result back to Y using 'sty'. The session also includes a practical example where the input data '02' in hexadecimal is multiplied to produce '08', with a step-by-step walkthrough of the emulator setup and execution.

05:04

📚 Practical Execution of Assembly Program in Emulator

The second paragraph details the practical execution of the assembly program in an emulator. It starts by setting up the input data '02' at memory location 4400 and then proceeds to run the program, which is pre-written in the editor. The program is assembled and loaded, and upon execution, the expected result '08' is observed at memory location 4410. The accumulator also reflects the same data, confirming the correctness of the operation. The paragraph concludes with a recap of the session's content and an anticipation for the next session, where more logical instructions will be covered, inviting viewers to join for further learning.

Keywords

💡8085 Assembly Language

The 8085 Assembly Language is a low-level programming language used to write programs for the Intel 8085 microprocessor. It is characterized by its use of mnemonics to represent machine code instructions. In the video, the language is used to write a program that manipulates data within the microprocessor's memory, demonstrating its role in executing low-level operations such as data multiplication.

💡Memory Location

A memory location in the context of the video refers to a specific address in the computer's memory where data or instructions are stored. The script mentions memory locations 'X' and 'Y', which are used to hold input data and the result of the operation, respectively. This concept is fundamental to understanding how data is accessed and manipulated in assembly language programming.

💡Multiplication by Four

Multiplication by four is the arithmetic operation performed in the video's program, where the data byte fetched from memory location 'X' is multiplied by four. This operation is achieved through a series of shifts, which is a common technique in assembly language to perform multiplication without using a dedicated multiplication instruction.

💡RLC Instruction

The RLC (Rotate Left Circular) instruction is a command in the 8085 Assembly Language that rotates the bits of the accumulator to the left. In the video, it is used twice to effectively multiply the accumulator's content by four, as each rotation left is equivalent to multiplying by two.

💡Accumulator

The accumulator is a register in the 8085 microprocessor used for arithmetic and logical operations. In the script, the accumulator is where the data from memory location 'X' is loaded and then manipulated using the RLC instruction to perform the desired multiplication.

💡Data Transfer Instruction

Data transfer instructions, such as 'ldax' and 'sty' mentioned in the video, are used to move data between memory locations and the accumulator. 'ldax' loads data into the accumulator, and 'sty' stores the accumulator's content into a specified memory location, which is essential for both input and output operations in the program.

💡HLT Instruction

The HLT (Halt) instruction is used to signal the end of a program in the 8085 Assembly Language. In the video, it is mentioned as a necessary component to indicate to the microprocessor that the program has concluded its execution.

💡Emulator

An emulator is a software that mimics the functionality of one system on another. In the context of the video, the emulator is used to run the 8085 Assembly Language program, allowing the viewer to see the program's execution and results in a simulated environment.

💡Input Data

Input data refers to the values provided to a program that it uses for processing. In the video, the input data is a hexadecimal value '02' stored at memory location '4400', which the program uses to perform the multiplication operation.

💡Output Data

Output data is the result produced by a program after processing the input data. The video demonstrates how the initial input data of '02' is multiplied by four, resulting in output data of '08', which is stored in memory location '4410'.

💡Logical Instructions

Logical instructions in assembly language are used to perform logical operations such as AND, OR, and NOT. Although not explicitly detailed in the script, the video mentions that the next session will cover problems involving a broader set of logical instructions, indicating their importance in assembly language programming.

Highlights

Introduction to the session on solving problems based on logical group of instructions.

Continuation from the previous session with a focus on 8085 assembly language programming.

Objective to write a program that multiplies a byte of data from memory location X by four and stores the result at memory location Y.

Explanation of the RLC instruction for multiplication by two in the 8085 microprocessor.

Strategy to multiply data by four by executing the RLC instruction twice.

Use of the ldax instruction to load data from memory location X into the accumulator.

Execution of the RLC instruction to perform the multiplication by four.

Utilization of the sty instruction to store the result in memory location Y.

Inclusion of the HLT instruction to signal the end of the program.

Practical demonstration of the program in an emulator with specific memory locations X and Y.

Input data setup in memory location 4400 with the value 02.

Explanation of bit shifting resulting in the output data in hexadecimal.

Conversion of input and output data values from hexadecimal to decimal for clarity.

Verification of the program's output by checking memory location 4410 for the expected result.

Emphasizing the importance of understanding the emulator for practical application of the program.

Assembly and loading of the program into the emulator for execution.

Observation of the program's successful execution and the correct output in memory location 4410.

Conclusion of the session with a teaser for the next session involving more logical instructions.

Appreciation and applause from the audience, signaling the end of the informative session.

Transcripts

[Music]

hello everyone from the previous session

itself we have been solving the problems

which are based on logical group of

instructions today as well we are going

to continue that strick so welcome to

the session logical instructions solve

problem two without any further Ado

let's get to

learning coming to the topic that we are

going to cover in this session

just like the previous session today as

well we are going to solve a problem on

logical group of

instructions consider the question write

an 8085 assembly language program that

takes data from the memory location X

then multiply this byte that is the data

of one bite which we have just fased

from the memory location X by

four and then stores the result at the

memory location y so let's try to solve

it now in the question it is mentioned

we are supposed to take out the data

from the memory location X then we are

supposed to multiply this data bite by

four if you remember we learned about

the instruction type

RLC and we saw it can be used for

performing multiplication by two so

clearly executing the instruction RLC

will multiply the data within the

accumulator by two but in this case we

are supposed to multiply the data bite

by four now notice using RLC the 8085

microprocessor is supposed to rotate the

content of the accumulator towards the

left in other words execution of RLC

will shift the contents within the

accumulator towards the left by one bit

now what will happen if we execute the

RLC two times well as I told told you

earlier Shifting the bits towards the

left one time means multiplication by

two therefore shifting it twice will

mean multiplication by 2 into 2 that is

4 so all we have to do is execute the

RLC instruction

twice let's now try and write the

program if you notice we are supposed to

take the data bite from the memory

location X and in order to execute RLC

see the data should be inside the

accumulator so we will use the data

transfer instruction

ldax now the data which was inside the

location X has already been loaded into

the accumulator so all we have to do is

execute the RLC instruction twice which

will specify that the data has been

multiplied by four now after this has

been performed we are also supposed to

store the result at the memory location

Y and for that we are going to use the

data transfer instruction

sty that is we will store the content of

the accumulator which due to the

execution of the RLC instruction twice

has already been changed because we

multiplied the data by four so the data

now using this instruction will be

stored inside the memory location y now

in order to complete the program in the

previous session itself I told you we

also need to have the instruction HLT

which will specify the microprocessor

that this is the Endo of the

program now this is the conceptual

program let's do this in Practical mode

and for that say we are choosing our

memory location where X is

440 and we are selecting our memory

location y to be

4410 let's now talk about the input data

save within the memory location

4400 we will store the input data 6

zeros followed by 1

0 now notice on this data we will

perform

RLC so once the data is loaded inside

the accumulator then if we execute the

instruction RLC once the bit one will

come to this place but we are not

executing the RLC instruction once we

are executing it

twice so the bit one will come to this

place so in the output data this will be

our bit sequence isn't it now in hexad

decimal can you give me the values of

both the input data and the output

data notice 0 0 1 0 is 2 and four zos is

0 so the input data in hexadecimal is

02 on the other

hand one triple 0 is 8 and four zos is 0

so this is going to be our output data

08 and that's pretty evident because if

you multiply two with four you are

supposed to get the value eight correct

So within the memory location

44 if we load

02 after the execution of this program

where we definitely will have to update

the memory locations X and Y

respectively we in the memory location

4410 we are supposed to get the data

08 let's now execute this program in the

emulator now in the previous session

itself we had a detailed walk through of

this emulator I encourage you to go

through the previous session so that you

can operate in this

emulator so as you can notice within the

editor I already have written down the

program for you this is the input data

location which was specified in the

question as X and this is the output

data location which was specified in the

program as y now like we discussed at

the memory location

440 we are going to update the input

data

02 so let's go to the memory location

44 so this is our memory location where

we will have our input data let's now

load the data we are supposed to load 0

2 so now in the memory location

4400 we have got the data

02 so both our input data and the

programs are ready let's now assemble

and load the program now the program has

been loaded all we have to do is run the

program notice as expected in the memory

location

4410 we have got the multiplier data

08 and since we executed the instruction

sta

4410 so the same data is also there

within the

accumulator so in this session we

covered the topic solve problem on

logical instructions we wrote another

program all right people that will be

all for this session in the next session

we are going to solve another problem

which is going to include most of the

logical instructions so I hope to see

you in the next one thank you all for

watching watching

[Applause]

[Music]