Program in 8085 Microprocessor

Question

There are N bytes stored from m/m location 2500H. The value of N is stored in 2400H.How can I write an 8085 program to interchange (irrespective of the bit value) the bit Di with Dj for all bytes. The values of i=4 and j=0

Answer 1

Nobody wants to do your homework for you. That being said, here's how you can interchange i-th bit of an 1-byte data with the j-th bit.

First let's revisit the bitwise logical operators and their usages. Suppose we want to know whether the 4-th bit is set, we take a bit mask 0000 1000 (i.e. 08H) and AND it with the data. To clear the 2nd bit we take the bit mask 1111 1101 (i.e. FDH) and AND it with data. Whereas, to set the 6-th bit we take a bit mask 0000 0010 (i.e. 02H) and OR it with the data. To complement of flip the 4-th bit we take a bit mask 0000 1000 (i.e. 08H) and XOR it with the data.

Assuming the actual data is in register D, thus to exchange the 2nd bit with the 4-th one we may write:

MVI A, 08H    ;i-th bit
ORI 02H       ;j-th bit
ANI D         ;only 2nd and 4bit of the data survives
JPE SKIP      ;if both bits are same (both 0 or both 1) no exchange required
;if not we need a swap, which is this case can be done by flipping the both 
MOV A,D       ;bring back the data again
XRI 08H       ;flip the i-th bit
XRI 02H       ;flip the j-th bit
SKIP: MOV D,A ;put the data back to D

The bit mask canbe programmatically generated with appropritate number of shifts (or rotate without carry). And repeating this process on all of the n bytes would complete the required task.