Virtual Gate Test Series: CO & Architecture - 8085 Microprocessor

Question

Consider the following instructions of an 8085 microprocessor
MVI D, 6 EH
MVI E, 5 DH
MOV A, D
ADD E
If above sequence of instructions are executed, then the value of carry flag (CY) and auxiliary carry (AC) flag respectively will be

1. CY = 0, AC = 0
2. CY = 0, AC = 1
3. CY = 1, AC = 1
4. CY = 1, AC = 0

Answer 1

Auxillary flag is set when we have carry out from the lower nibble to higher nibble..And carry flag is set when carry out from MSB..So we add the binary equivalent of the given hexadecimal numbers.

     0110  1110

+   0101  1101

---------------------

     1100   1011

We have done this addition keeping in mind the carries..

So here clearly there is no carry out of MSB but carry out from lower nibble to higher nibble..

So , CY =  0  , AC = 1

Hence B) option is correct..

Comments

if instruction is MVI D,8ABC H
then value using little endian will be BC8A
am i correct ?
please reply

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little endian means lower byte have lower address and higher byte have higher address.

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that means value will be 8ABC but its storage in memory will be different using little endian ?