Word 20 contains 40 Word 30 contains 50 Word 40 contains 60 Word 50 contains 70 Which of the following instructions loads 60 into the accumulator ? (a) load immediate 60 (b) load direct 30 (c) load indirect 20 (d) load indirect 30 Which of the above statements are correct ?
(a) and (c)
The three addressing modes, applied to this memory map.
| Address | Contents |
|---|---|
| 20 | 40 |
| 30 | 50 |
| 40 | 60 |
| 50 | 70 |
(a) load immediate 60. The operand is the value: 60 goes straight into the accumulator. No memory is read at all. ✓ loads 60.
(b) load direct 30. The operand is an address; the CPU fetches whatever that location holds. Word 30 contains 50, so the accumulator gets 50. ✗
(c) load indirect 20. Indirect addressing takes two memory accesses. The operand 20 is the address of a pointer: first read word 20, which contains 40, and treat that as the real address; then read word 40, which contains 60. ✓ loads 60.
\(20 \rightarrow 40 \rightarrow 60\)
(d) load indirect 30. Following the same two hops: word 30 contains 50, so go to word 50, which contains 70. The accumulator gets 70. ✗
So (a) and (c) both deliver 60, which is option 1.
The general rule worth fixing. Each addressing mode adds one level of indirection:
| Mode | What the operand is | Memory reads |
|---|---|---|
| Immediate | the data itself | 0 |
| Direct | the address of the data | 1 |
| Indirect | the address of the address | 2 |
Why indirect addressing exists. Because the pointer sits in memory, it can be changed at run time — which is exactly how arrays are stepped through, how linked lists are followed, and how parameters are passed by reference. The cost is the extra memory access, which makes it the slowest of the three modes.
Hence, the correct answer is (a) and (c).
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ADD C
DAA
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A single instruction to clear the lower four bits of the accumulator in 8085 assembly language is-
In Microprocessor 8085 Address/Data buffer is a/an _______ buffer.
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