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8086 Data Addressing Modes


Posted by: seminar class
Created at: Tuesday 15th of February 2011 04:21:45 AM
Last Edited Or Replied at :Tuesday 15th of February 2011 04:21:45 AM
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diate Addressing ModeDirect Addressing ModeRegister Addressing Mode
8086 processor has some additional features like:
Additional registers, Extra Segment (ES) and Stack Segment (SS)
Addressable physical memory of 1Mb (8085 has 64Kb)
These features have an impact on the addressing modes of 8086. The
general addressing modes of 8086 processor therefore can be classified as:Data Addressing Modes – This mode is related to data transfer operations,that is, data is transferred either from the memory to internal registers of8086 processor or from one register to another register.
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Created at: Thursday 18th of October 2012 03:45:20 AM
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Created at: Sunday 07th of October 2012 12:45:39 PM
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8086 Data Addressing Modes


Posted by: seminar class
Created at: Tuesday 15th of February 2011 04:21:45 AM
Last Edited Or Replied at :Tuesday 15th of February 2011 04:21:45 AM
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ction, the code execution control jumps to
the current code segment location addressed by the contents of AX
register.
Stack Memory Addressing Modes – This mode involves stack registry
operations.
Example: PUSH AX, this instruction copies the contents of AX register to thestack.In this session, we will focus on the data addressing modes of 8086. Programand Stack memory addressing mode will be discussed in the next session.
Data Addressing Modes of 8086 Processor

The additional data addressing modes in 8086 are:
1. Register Indirect Addressing Mode
2. Base Relative-Plus-Inde..................[:=> Show Contents <=:]



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Posted by: computer science technology
Created at: Sunday 24th of January 2010 10:41:31 AM
Last Edited Or Replied at :Friday 23rd of November 2012 12:02:15 AM
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ious value of B and B is set equal to the value of A mod B.
similarly, at each step of extended Euclidâ„¢s algorithm,A3 is set equal to the previous value of B3,and B3 is set equal to the previous value of A3 minus the integer quotient of A3/B3 multiplied by B3.This latter value is simply the remainder of A3 divided by B3, which is A3mod B3.
Note also that if gcd (m, b) =1, then on the final step we would have B3=0 and A3 =1. Therefore, on the preceding step, B3 =1. But B3 =1, then we can say the following:

mB1+bB2=B3
mB1+bB2=1
bB2=1-mB1
bB2= 1mod m
And B2 is the multiplicat..................[:=> Show Contents <=:]



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