Assembly language to Machine Code Conversion in 8086 || Basic Understanding of Instruction format

[@vishaldemo]

Brilliant explanation of all the questions, thanku sir😍😍

[@LearnOnline1]

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[@taqqiraja2722]

You're making Pakistan proud

[@LearnOnline1]

Jazak Allah. I am glad to hear this comment. :) Its my utmost desire.

[@deedewald1707]

Brilliant explanation and process ! It's now so easy to flow the procedures !

[@LearnOnline1]

Thanks !!

[@mememaster9876]

Thank you so much for this video

[@LearnOnline1]

You are so welcome!

[@legacywolf443]

Thanks, helped a lot, it's what I've been searching for! Very underrated video.

[@LearnOnline1]

You are welcome. I am pleased to know that It was useful video for someone to learn

[@meharhijjazi8281]

Brilliant explanation

[@LearnOnline1]

Glad you liked it

[@renukadevi5250]

Clear explanation... ADD [BX+SI], AL CONVERT INTO MACHINE LANGUAGE WHERE SI=2038, MOV [EBX+4*ECX+EF34AB62H], 7FAC88BDH convert into machine language...thank u in advance

[@Araf38]

That was great sir.. Take love from Bangladesh ❣️

[@LearnOnline1]

Thankyou so much for your regards!! I am pleased that you liked it!

[@twmarufu]

Love from Namibia

[@LearnOnline1]

Jazak Allah!! Thank you. I am pleased that my little effort was useful for you.

[@rashidyaseen6270]

what is the answer of add [5024H], DX

[@engr.ayazsoaringthoughts]

Brilliant logics sir

[@LearnOnline1]

Thankyou !!

[@dipsiecamila]

Do you have the same video as X80_64, or does one else have a video of how to encode 64 bits? Please...

[@farehaamir7438]

Assalam o alaikum sir. Thank you so much sir. Sir may you plz share the answers of example with us?

[@LearnOnline1]

Wailekum salam, I did not understand your question. What examples are you talking about?

[@kMuhammadRayanAli]

great video. RESPECT++

[@LearnOnline1]

Thanks !! I appreciate your encouraging words!

[@deedewald1707]

Yes, very clear !

[@LearnOnline1]

Thanks. I am happy it was useful for you.

[@muneebkhalil5958]

great sir g

[@pablodinannounsada]

Hi Sir, do you have the complete mnemonic set of the 8086? thanks and regards

[@LearnOnline1]

If you meant the instruction set. it is available on google for free. I hope you can simply find it on the first page.

[@deedewald1707]

I like your original comment !

[@yassenramadan9069]

Sir in ex 1 you consider Al as a register and in ex 2 you consider Cx as a register why ?

[@LearnOnline1]

if you both are register, you can choose any of the register as REG and R/M but it is highly recommended to choose Destination Operand as REG so there are two approaches.. Thats what i have done in exercise 1, . I Have used both approaches. In first approach, MOV BL, AL. I kept AL as source operand (AL) as REG . But as per recommendation in second approach So BL (destination operand) as REG. While in exercise 2, I have used only second approach which is highly recommended . MOV CX,DX So CX(Destination operand) as REG. I hope , I have clarified the your confusion.

[@yassenramadan9069]

@@LearnOnline1 thanks sir ❤️❤️

[@LearnOnline1]

@@yassenramadan9069 You are welcome

[@wassup102]

Thank you so much sir

[@LearnOnline1]

You are welcome!!

[@AbbasAli-zy1po]

a) JMP AX b) JMP LIST[DX] c) JMP NEAR PTR[DI+3] indirect program memory

[@LearnOnline1]

I If you want to understand these instructions.. Please follow this tutorial kzbin.info/www/bejne/aJTVnJKfh7SHZ6M

[@hammadjadoon6347]

awesome

[@LearnOnline1]

Thanks! I am pleased to know that my content was useful.

[@luisgeneroso4151]

I would to meet with you face time, if you have time. looking for more in depth info, regarding your acknowledgement on this concept. How can i reach you?

[@LearnOnline1]

Email me at [e-mail]

[@hamzashakeel7661]

Can you help me with this? I need machine code of it Where the OPCODE of MOV is 100010 a. MOV ARR, DI b. MOV [BP+DI+96h], BL c. MOV [BX+DI+100h], 2019h d. MOV DX , [DI] e. MOV [BX], SI

[@LearnOnline1]

I believe examples on such samples are already solved in this video tutorial.

[@無心插柳柳橙汁-k5h]

sir thank you so much

[@LearnOnline1]

You are welcome. I am happy to know that it was useful.

[@saqibmuhammad8614]

Well done sir

[@LearnOnline1]

Thank you

[@saqibmuhammad8614]

Salam

[@saqibmuhammad8614]

Sir apka WhatsApp number he

[@LearnOnline1]

@@saqibmuhammad8614 wailekum salam

[@LearnOnline1]

@@saqibmuhammad8614 you can share your comments here or if you have something to attach then you can email me at [e-mail]

[@muhaiminahmad7953]

Walaikum salam. Jazakallah khoiran

[@LearnOnline1]

Ameen!!

[@fariajitu9213]

What does actually mean by displacement?

[@LearnOnline1]

Displacement is numerical address present with inexed or based registers.

[@faranak4355]

thanks :)

[@LearnOnline1]

You are welcome. I am pleased it was useful for you.

[@minzabatool8529]

How get to know about 8 bit or 16 bit

[@LearnOnline1]

If you are talking about instruction size bit that's normally referred as W bit in Assembly to machine code. then it is determined with the help of Register size . Let say your instruction isn MOV AX, [SOME MEMORY] SO here register is AX and it is a 16-bit so w bit will be 16-bit. For your convenience just keep in mind that AX, BX, CX and DX are 16-bit. While AL, AH, BL, BH, CL, CH, DH and DL are 8-bit registers. I Suggest you go through my data registers of 8086 or architectures of 8086. These are one of the basic and fundamental tutorials of this tutorial series.

[@obinashchandrashil3761]

1) MOV BX, SS: [BP+SI] (2) MOV [EBX+4*ECX] can you conver this to machine language pls?

[@LearnOnline1]

First Instruction , you asked about. It uses something called segment override prefix. To encode the assembly language of such instruction you need to use one extra byte before the general format of instruction. Format would be override byte , General format given in this format. override byte format = 001 SEG 110,,SEG is 2-bit field which could be any of ES, CS, SS and DS and their respective values will be 00, 01,10,11. So According to Instruction MOV BX, SS: [BP][SI] Over ride byte= 001 10 110= 36 Hex Opcode =100010 d= 1 w=1, mod=00,reg=011,r/m =010 (Well explained in the above video) 10001011=8B HEX, 00011010=1A Hex so the over all mahine code would be 36 8B 1A hex (3 bytes instruction) Second Instruction,: this is not instruction which you can encode using the guidelines in given video which is about 8086 microprocessor. While the instruction you mentioned is for x8086. So I suggest to read some book or go through the video tutorial about X8086. I have also noted it. Soon, I will also try to make tutorial about x8086 and post in this channel.

[@KenKaneki-oy3kb]

what if its a 32bit ? what code for it?

[@LearnOnline1]

If you meant 32-bit microprocessor of x86. Then it will have some other extra rules following this tutorial. But, if you meant 32-bit number then this tutorial is valid.

[@nforion6129]

what is the ans of MOV [BP][SI] + 24H, BX

[@LearnOnline1]

I am sorry for such late response. I was extremely busy . I hope it still have some worth. Opcode =100010, D=0,W=1 ; Byte1= 10001001=89H REG=011, R/M=010, MOD=01; BYTE2= MOD REG R/M =01 011 010 =0101 1010 = 5A hex Byte 3 is 8 bit displace given in instruction (24H ) so the answer would be 89 5A 24 H

[@rezaprantick7380]

MOV AH, 4CH what will be machine code

[@LearnOnline1]

I believe you have written an incorrect instruction. MOV AH,CH OPCODE, D,W MOD REG R/M OPCODE =100010 D=0 , REG= BX=100, W=0 MOD=11, R/M=CH=101 100010 0 0 11 100 101 1000 1000 1110 0101 =88E5H

[@rezaprantick7380]

@@LearnOnline1 what if MOV AH, 4??

[@LearnOnline1]

@@rezaprantick7380 Kindly watch this tutorial I have made this specific video tutorial for your this specific question. kzbin.info/www/bejne/iqKqfmOuhNWjf5o

[@deedewald1707]

MOV AH,4Chex I got B4 4C hex ! MOV AX,1234hex I get B8 34 12 hex !

[@deedewald1707]

@@LearnOnline1 the request is mov AH,4Chex I got B4 4C hex !

[@Tarun-hj1hs]

Subtraction?

[@LearnOnline1]

What do you mean by Subtraction?

[@deedewald1707]

MOV is 1000 10 ADD is 0000 00 SUB Opcode is ?

[@Trilogygaming123]

Sir i want admission in cui

[@LearnOnline1]

You can contact CUI Admission offices.