Assembly Language Programming Questions and Answers

Q: Match List I with List II and select the correct answer using the codes given below :List IList II(a) Register addressing(i) MOVC A, @ A + PC(b) Immediate addressing(ii) DJNZ R2, LOOP(c) Relative addressing(iii) MOV A, # 55H(d) Indexed addressing(iv) MOV A, R1

The correct code is (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).MOV A, R1 takes its operand from a register, which is register addressing. MOV A, #55H carries the data 55H in the instruction itself, immediate addressing. DJNZ R2, LOOP jumps by a signed offset relative to the program counter, relative addressing. MOVC A, @A+PC forms the address by adding the accumulator to a base register, indexed addressing, used for look-up tables in code memory.The other codes give immediate addressing to MOV A, R1 or treat the DJNZ jump as indexed.

Q: The JS is called as ...............

JS stands for jump if sign (the 'jump the signed bit' option).In the 8086 instruction set, JS is a conditional jump taken when the sign flag SF is 1, that is, when the result of the last arithmetic or logic operation was negative (its most significant bit is 1). Its complement is JNS, jump if not sign.'Jump single bit', 'jump simple bit' and 'jump signal it' do not describe any flag. The S in JS refers to the sign flag, so the option mentioning the signed bit is the one that fits.

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Directions : Q. Nos. 21 to 30 : The following items consist of two statements, one labelled the “Assertion (A)” and the other labelled the “Reason (R)”. You are to examine these two statements carefully and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer accordingly.

Assertion (A) : The top down structured programming should be used for developing programmes.

Reason (R) : The top down structured programming methodology enables us to get readable and easily provable program.

A.Both (A) and (R) are true and (R) is the correct explanation of (A).
B.Both (A) and (R) are true, but (R) is not correct explanation of (A).
C.(A) is true, but (R) is false.
D.(A) is false, but (R) is true.

Exam Year(s)

UGC NET - 2012,

Answer: A

The keyed answer is that both statements are true and (R) is the correct explanation of (A). In top-down structured programming a problem is broken into smaller modules, each handled in turn, so the program is easier to read, test and prove correct. That benefit is the reason why the method should be used for developing programs. Option B, which says that (R) does not explain (A), is rejected by the key, and options C and D would call one of two accepted statements false.

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Match List I with List II and select the correct answer using the codes given below :

List I	List II
(a) Register addressing	(i) MOVC A, @ A + PC
(b) Immediate addressing	(ii) DJNZ R2, LOOP
(c) Relative addressing	(iii) MOV A, # 55H
(d) Indexed addressing	(iv) MOV A, R1

A.(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
B.(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
C.(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)
D.(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The correct code is (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).

MOV A, R1 takes its operand from a register, which is register addressing. MOV A, #55H carries the data 55H in the instruction itself, immediate addressing. DJNZ R2, LOOP jumps by a signed offset relative to the program counter, relative addressing. MOVC A, @A+PC forms the address by adding the accumulator to a base register, indexed addressing, used for look-up tables in code memory.

The other codes give immediate addressing to MOV A, R1 or treat the DJNZ jump as indexed.

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The JS is called as ...............

A.jump the signed bit
B.jump single bit
C.jump simple bit
D.jump signal it

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

JS stands for jump if sign (the 'jump the signed bit' option).

In the 8086 instruction set, JS is a conditional jump taken when the sign flag SF is 1, that is, when the result of the last arithmetic or logic operation was negative (its most significant bit is 1). Its complement is JNS, jump if not sign.

'Jump single bit', 'jump simple bit' and 'jump signal it' do not describe any flag. The S in JS refers to the sign flag, so the option mentioning the signed bit is the one that fits.

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In the Assembly program, the following steps are to be followed. Find the sequence in which the program to be written for 8085 microprocessor.

(i) Initialization of variables.
(ii) Initialization of stack.
(iii) Enable or disable interrupts.
(iv) Program should be completed with last line as 'END' statement.

A.(i), (iii), (ii) and (iv)
B.(i), (ii), (iii) and (iv)
C.(ii), (i), (iii) and (iv)
D.(ii), (iii), (i) and (iv)

Exam Year(s)

UGC NET - 2013,

Answer: C

The usual sequence in an 8085 program is the initialisation of the stack, the initialisation of variables, the enabling or disabling of interrupts, and then the program body ending with the END statement, which is (ii), (i), (iii), (iv), the third option. The stack pointer has to be set first so that calls and interrupts have a place to save the return address.

The first two options start with the variables before the stack, and the fourth puts the interrupts before the variables, so they do not follow this order.

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Assembly Language Programming Questions and Answers

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