Logical Instructions in 8085 Microprocessor (Solved Problem 2)
Summary
TLDRIn 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
🤖 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.
📚 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.
Mindmap
Keywords
💡8085 Assembly Language
💡Memory Location
💡Multiplication by Four
💡RLC Instruction
💡Accumulator
💡Data Transfer Instruction
💡HLT Instruction
💡Emulator
💡Input Data
💡Output Data
💡Logical Instructions
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]
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