A single instruction to clear the lower four bits of the accumulator in 8085 assembly language is-
ANI F0H
The question asks for a single instruction in 8085 assembly language to clear the lower four bits of the accumulator. Clearing bits means setting their values to 0. We need an instruction that performs a logical operation between the accumulator's current value and a specific mask, such that the result in the accumulator has the lower four bits as 0, while the upper four bits remain unchanged.
In 8085 assembly, logical instructions like AND, OR, and XOR are commonly used for bit manipulation with an immediate value. Let's look at the relevant instructions provided in the options:
Based on this, the ANI instruction is suitable for clearing specific bits while preserving others by using an appropriate mask.
We want to clear the lower four bits (bits 0, 1, 2, 3) of the accumulator and keep the upper four bits (bits 4, 5, 6, 7) unchanged. Using the ANI instruction, the immediate data acts as a mask:
So, the required 8-bit mask in binary form is:
1 1 1 1 0 0 0 0
This mask has 1s for the upper four bits (to preserve them) and 0s for the lower four bits (to clear them).
Let's convert this binary mask to hexadecimal:
So, the hexadecimal mask is F0H.
Now let's examine the options with the mask F0H:
ANI 0FH: The mask is 0000 1111. This would clear the upper four bits and preserve the lower four bits.
XRI 0FH: The mask is 0000 1111. This would invert the lower four bits and preserve the upper four bits.
XRI F0H: The mask is 1111 0000. This would invert the upper four bits and preserve the lower four bits.
ANI F0H: The mask is 1111 0000. This would preserve the upper four bits (AND with 1) and clear the lower four bits (AND with 0). This matches our requirement.
Let's illustrate the operation of ANI F0H with an example. Assume the accumulator (A) contains the value A5H (binary 1010 0101).
| Bit Position | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|---|---|---|
| Accumulator (A) | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 |
| Mask (F0H) | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
| Result (A AND F0H) | 1&1=1 | 0&1=0 | 1&1=1 | 0&1=0 | 0&0=0 | 1&0=0 | 0&0=0 | 1&0=0 |
The result in the accumulator is 1010 0000, which is A0H. As you can see, the upper four bits (1010) remain unchanged, and the lower four bits (0101) are cleared to 0000.
The instruction that performs a bitwise AND operation between the accumulator and the mask F0H successfully clears the lower four bits while keeping the upper four bits intact. This instruction is ANI F0H.
| Instruction | Operation | Immediate Data (Mask) | Effect on Bits where Mask bit is 1 | Effect on Bits where Mask bit is 0 |
|---|---|---|---|---|
| ANI data | Accumulator = Accumulator AND data | Any 8-bit value | Preserve original bit (A & 1 = A) | Clear bit (A & 0 = 0) |
| ORI data | Accumulator = Accumulator OR data | Any 8-bit value | Set bit (A | 1 = 1) | Preserve original bit (A | 0 = A) |
| XRI data | Accumulator = Accumulator XOR data | Any 8-bit value | Invert bit (A ⊕ 1 = ¯A) | Preserve original bit (A ⊕ 0 = A) |
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