Intel 8086 and 8088

segment is for converting physical address to logical address , here on taking 8086 microprocessor as example, we have 20 address lines but it is capable of taking only 16 address lines.... so to convert that 20 into 16 segment is used....

Minimum mode describes a chip configuration where the 8288 bus controller is not required. Certain signals, such as ALE, are then generated directly by the processor. Maximum mode describes a chip configuration where the 8288 bus controller is required. The 8288 generates ALE, and other signals, allowing the processor to provide other signals, such as LOCK. These two mode allow the system designer some freedom, and allow the number of pins on the chip to remain lower.

The 8086 has 16 data bus lines and 20 address bus lines because that is how Intel designed it. They wanted a processor that was more powerful than the 8085, which has an 8 bit data bus and a 16 bit data bus, so they increased both bus sizes accordingly.

BHE is used to enable the high order bus so as to differentiate between a word operation and a byte operation.

During a single bus cycle, the 8-bit microprocessor transfers one byte while the 16-bit microprocessor transfers two bytes. The 16-bit microprocessor has twice the data transfer rate.

The response to another interrupt request during servicing of an interrupt depends on the setting of the interrupt enable flag and/or the interrupt mask. If the interrupt service routine is thread safe, it will process correctly. More probably, problems will arise so the proper procedure is to either set the interrupt mask (8085) to not allow this or lower priority interrupts before reenabling interrupt, or to leave interrupts disabled until this iteration is complete.

If interrupts are disabled or masked, a subsequent interrupt will be posted but not processed until the first is completed.

An instruction queue is used in the 8086 to speed up the average time it takes to process an instruction. Some instructions are faster than the bus, while some are slower. If the CPU had to wait for all of the instructions, there would be gaps of time where the CPU is doing nothing. The queue helps to eliminate that gap by prefetching instructions in the hope that they will be ready for use when the CPU gets to them.

interrupt vector table

RIM is Read Interrupt Mask. It is used to access the interrupt mask register, which contains the status of the three interrupt masks and the serial input data pin.

One many find this answer on YouTube. One also may find out how to write ascending order programs using an 8086 microprocessor by looking at the owners manual.

Physical address in the 8086/8088 is {Selected Segment Register} * 16 + {Effective Offset Address}. It is a 20-bit address

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yes it s rude and so annoying

In 8086 pipeline concept was introduced bcoz in 8086 we started using a buffer space which fetches the instruction at the same time when CPU processes

Near calls and returns transfer control between procedures in the same code segment. Far calls and returns pass control between different segments.

mov ax, @data ; Initialize data section

mov ds, ax

mov bx, offset lookup ; Load offset of lookup in bx

mov al, key ; key no. in al

xlat ; translate byte in al

mov bh, al ; al = lookup(key)

mov ch, 02h ; Count of digits to be displayed

mov cl, 04h ; Count to roll by 4 bits

l2: rol bh, cl ; roll bl so that msb comes to lsb

mov dl, bh ; load dl with data to be displayed

and dl, 0fH ; get only lsb

cmp dl, 09 ; check if digit is 0-9 or letter A-F

jbe l4

add dl, 07 ; if letter add 37H else only add 30H

l4: add dl, 30H

mov ah, 02 ; Function 2 under INT 21H (Display character)

int 21H

dec ch ; Decrement Count

jnz l2

mov ah, 4cH ; Terminate Program

int 21H

end

8086 has memory divided into segments.
Each segment has its particular register like ES, DS, SS, CD (extra segment, data segment, stack segment and code segment).
These registers hold the base address BA.

Now, there two base registers(BX and BP) and two index registers(SI and DI) in 8086. These registers hols the effective address EA.

Now Physical address PA is sum of EA and BA.

That is,

PA = BA+EA

I believe a nested interrupt, is where an interrupt is allowed to occur (and thus is handled) during an already occurring Interupt service ruotine.

I.E.

First interrupt occurs

ISR1 begins

second Interrupt occurs

ISR2 begins

ISR2 Finishes

ISR1 continues from where left off

ISR1 finishes

http://infocenter.arm.com/help/index.jsp?topic=/com.arm.doc.dvi0022a/ar01s02s09.html

The i7 is a 64 bit processor.

it's instruction pointer register it's in cpu and it holds the instruction which the cpu fetching it from memory

Pins 1 and 20 in the 8086 microprocessor are (both) power and signal ground (GND).

it reduces internal fragmentation....on 8086 its limited to a max of just 16b per segment of each process....in other words if they weren't overlapping this would be 2^16!

btw I'd like to hear more possible explanations too!

If you mean greater (more) 8086.

The stack register contains the address of the last item pushed on the stack. The stack is a region of memory used for temporary storage of instruction addresses and register values in a Last-In-First-Out (nested) structure. It is used for interrupts and subroutine calls.