Each instruction in an assembly program has the following fields : (i) Lable field Please write the proper sequence of fields :
(ii) Operand field
(iii) Comment field
(iv) Mnemonic field
(i), (iv), (ii) & (iii)
An assembly statement reads left to right as label, mnemonic, operand, comment — (i), (iv), (ii), (iii), option 2.
LOOP: MVI A, 05H ; load 5 into A
| Field | In the example | Purpose | Optional? |
|---|---|---|---|
| (i) Label | LOOP: | Names the address, for jumps and calls | Yes |
| (iv) Mnemonic | MVI | The operation — assembles to the opcode | No |
| (ii) Operand | A, 05H | What the operation acts on | Depends |
| (iii) Comment | ; load 5 into A | Explanation for the reader | Yes |
The order is not arbitrary; each field's position follows from its job. The label must come first because it marks the address at which the instruction begins — it is a name for the location, so it stands before anything the assembler emits. The mnemonic must precede its operands because the operation determines how many operands are expected and how they are interpreted: MVI takes a register and a byte, INR takes a register alone, NOP takes none. And the comment must come last because the delimiter that introduces it — a semicolon in Intel syntax — tells the assembler to ignore the rest of the line. Put a comment anywhere else and everything after it vanishes.
Only the mnemonic is compulsory, which is worth remembering because it explains why options beginning with a comment are wrong in principle as well as in practice: a line consisting of a comment alone is legal, but it is not an instruction, and the question asks about the fields of an instruction.
How the assembler uses the labels is the reason two passes are needed: a forward jump refers to a label defined later in the file, so the first pass builds a symbol table of label addresses and the second pass generates the machine code with those addresses filled in.
The value of labels is that they survive editing. Writing JMP LOOP instead of JMP 2050H means that inserting an instruction earlier in the program does not silently break the jump — the assembler recomputes the address.
Hence, the proper sequence is (i), (iv), (ii) & (iii).
Similar Questions
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List – I List – II a. RLC i. Rotate Accumulator Right through carry b. RRC ii. Rotate Accumulator Left through carry c. RAR iii. Rotate Accumulator Left d. RAL iv. Rotate Accumulator Right Codes :
Match the following :
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i. 8279
ii. 8155
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Important Questions from Microprocessors
An 8086 address bus is a/an _____ -bit bus
Match the columns.
Pin
Description
a. D 0-D 7
1. Reset Input
b. RESET
2. Data Lines
c. A 0,A 1
3. Internal Address
An 8 bit microcontroller has an external RAM with the memory map from 8000 H to 9FFF H. How many number of bytes can this RAM store?
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Which of the following 8085 microprocessor instructions occupies 3 bytes in memory?
