All 4 segments are not necessary . Only code segment is the compulsory segment. Size of each segment cannot exceed 64kb. You can initialise each segment in any order . You can give spaces between segments so that you can modify your program and if it becomes longer does not over lap with another segment. You can have all four segments inside 64kb that mean all segments can overlap but as a programmer you have decide. Memory size of 1MB does not mean 64kb x 4 segments. Hope you got the answer.
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![[1.2] 8086 Microprocessor Architecture](/img/tr.png)
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Thank you so much , it was an easy and simple explanation. Would love to watch more of your content on 8086 microprocessor.
Finally smjh agya jiske liye itni videos dekhi
Finally it is cleared to me, thanks a lot for this easy explationation.
good job!
nice video , very helpfull
Nice job mate!
Very nice
Make more videos on 8086
Thanks for the motivation
Thank you very much
👍👍
✌️✌️✌️✌️
thankyou sooo much
I didn't understand the 1234h and 5678h !
The test question and one should calculate the physical address using the formula !
6:40 but 64kb + 64kb + 64kb + 64kb =/= 1Mb
All 4 segments are not necessary . Only code segment is the compulsory segment. Size of each segment cannot exceed 64kb. You can initialise each segment in any order . You can give spaces between segments so that you can modify your program and if it becomes longer does not over lap with another segment. You can have all four segments inside 64kb that mean all segments can overlap but as a programmer you have decide. Memory size of 1MB does not mean 64kb x 4 segments. Hope you got the answer.
Note: a com program has the 64kb sized limit for the entire project ! Executables have limit also !