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## 8086 Data Addressing ModesPosted 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 | addressing modes in 8086 microprocessor ppt ,
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diate Addressing ModeDirect Addressing ModeRegister Addressing Mode8086 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. Examp.................. [:=> Show Contents <=:] | |||

## strassen matrix multiplication examples pptPosted by: Created at: Thursday 18th of October 2012 03:45:20 AM Last Edited Or Replied at :Thursday 18th of October 2012 03:45:20 AM | discuss strassen s matrix multiplication ppt ,
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## 8086 Data Addressing ModesPosted 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 | addressing modes in 8086 microprocessor ppt,
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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 ProcessorThe additional data addressing modes in 8086 are: 1. Register Indirect Addressing Mode 2. Base Relative-Plus-Inde.................. [:=> Show Contents <=:] | |||

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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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