What idea does the program use to exchange bits D6 and D2?

Toggle both bits if they differ; otherwise leave the byte unchanged

Clear D6 and D2, then set them both to 1

Rotate the byte 4 times and store the rotated byte

Why is the byte rotated left four times (RLC x4) before XRA B?

To bring D2 to the D6 position, so XRA B gives D2 xor D6 at D6

To move D6 into the carry flag so that JC can test it

To swap the nibbles so the two bits are exchanged directly

To multiply the byte by 16 before masking with 40H

How does the program process all 200 bytes from 8090H?

D starts at 0C8H; each byte is written back in place, then INX H, DCR D

D starts at 200H and the results go to a new table

D starts at 0C8H and HL is decremented after each byte

It loops until it finds a 00H byte in the table

2070 Bhadra · 8 marks

8085 ALP: exchange bits D6 and D2

Short answer

For each byte, D6 and D2 are swapped in place. Rotate left 4 times so D2 moves to D6; XOR with original isolates D6 xor D2 at 40H, mask it, shift to D2 (04H), OR to get mask 44H if bits differ, else 00H. XOR with original toggles both bits, exchanging them. Processed 200 bytes.

Write an assembly language program for 8085 to exchange the bits D6 and D2 of every byte of a program. Suppose there are 200 bytes in the program starting from memory location 8090H.

Assumptions: the 200 bytes (C8H) start at 8090H and are exchanged in place.

Method: D2 and D6 are 4 positions apart. If the two bits are different, toggling both exchanges them; if equal, nothing changes. The byte is rotated left 4 times so D2 is in the D6 position, XOR with the original gives D2 xor D6 at D6, which is masked with 40H; copying it 4 places right gives the other bit position (04H), and OR gives the toggle mask 44H or 00H.

        LXI H,8090H     ; HL -> first byte of the program area
        MVI D,0C8H      ; counter = 200 decimal
LOOP:   MOV A,M         ; A = byte
        MOV B,A         ; save original byte
        RLC             ; rotate left 4 times: D2 comes to D6
        RLC
        RLC
        RLC
        XRA B           ; D6 of A = D2 xor D6 of original
        ANI 40H         ; 40H if D2 and D6 differ, else 00H
        MOV C,A
        RRC             ; 40H -> 04H (position D2)
        RRC
        RRC
        RRC
        ORA C           ; 44H if they differ, else 00H
        XRA B           ; toggling both bits = exchanging them
        MOV M,A         ; write back in the same place
        INX H
        DCR D
        JNZ LOOP
        HLT

Check: byte 40H (D6 = 1, D2 = 0) gives mask 44H, so 40H xor 44H = 04H (bits exchanged).

✓ Answer checked by Claude Opus 5.5 (high effort)

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