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Test: Jump, Loop And Call Instructions - Computer Science Engineering (CSE) MCQ


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10 Questions MCQ Test - Test: Jump, Loop And Call Instructions

Test: Jump, Loop And Call Instructions for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Test: Jump, Loop And Call Instructions questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The Test: Jump, Loop And Call Instructions MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Jump, Loop And Call Instructions below.
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Test: Jump, Loop And Call Instructions - Question 1

DJNZ R0,label is how many bit instruction?

Detailed Solution for Test: Jump, Loop And Call Instructions - Question 1

DJNZ R0,label has hex code, D8 and its corresponding label address. Hence, it is a two byte instruction.

Test: Jump, Loop And Call Instructions - Question 2

JZ, JNZ, DJNZ, JC, JNC instructions monitor the bits of which register ?

Detailed Solution for Test: Jump, Loop And Call Instructions - Question 2

PSW register consists of flag bits like CY, P, Z etc so it is the one that is directly being monitored by these instructions.

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Test: Jump, Loop And Call Instructions - Question 3

Calculate the jump code for again and here if code starts at 0000H​
MOV R1,#0
MOV A,#0
MOV R0,#25H
AGAIN:ADD A,#0ECH
JNC HERE
HERE: INC R1
DJNZ R0,AGAIN
MOV R0,A
END

Detailed Solution for Test: Jump, Loop And Call Instructions - Question 3

Loop address is calculated by subtracting destination address and the address next to the source address.
So here if we start with 0000H
then source address is 0008H and destination address is 0004H
So loop address is 04-0A=E9

Test: Jump, Loop And Call Instructions - Question 4

 When the call instruction is executed the top most element of stack comes out to be

Detailed Solution for Test: Jump, Loop And Call Instructions - Question 4

topmost element of the stack is the address of the instruction next to the call instruction so that when RET is executed then PC is filled with that address and so the pointer moves to the main program and continue with its routine task.

Test: Jump, Loop And Call Instructions - Question 5

 LCALL instruction takes

Detailed Solution for Test: Jump, Loop And Call Instructions - Question 5

LCALL instruction moves the pointer to a 16 bit address so it is a 3 byte instruction.

Test: Jump, Loop And Call Instructions - Question 6

 Are PUSH and POP instructions are a type of CALL instructions?

Detailed Solution for Test: Jump, Loop And Call Instructions - Question 6

PUSH and POP instructions are not CALL instructions because in POP and PUSH instructions the pointer does not move to any location specified by its address which is the fundamental of CALL instruction, so it is not a type of CALL instruction.

Test: Jump, Loop And Call Instructions - Question 7

 What is the time taken by one machine cycle if crystal frequency is 20MHz?

Detailed Solution for Test: Jump, Loop And Call Instructions - Question 7

Time taken by one machine cycle is calculated by the inverse of a ( crytal frequency ) /12

Test: Jump, Loop And Call Instructions - Question 8

 Find the number of times the following loop will be executed

MOV R6,#200
BACK:MOV R5,#100
HERE:DJNZ R5, HERE
DJNZ R6,BACK
END

Detailed Solution for Test: Jump, Loop And Call Instructions - Question 8

 It will be executed 200*100 times.

Test: Jump, Loop And Call Instructions - Question 9

What is the meaning of the instruction MOV A,05H ?

Detailed Solution for Test: Jump, Loop And Call Instructions - Question 9

 If we need to store the address in the accumulator, then directly the address is moved to it unlikely of using # used for storing data in any register.

Test: Jump, Loop And Call Instructions - Question 10

 Does the two instructions mean the same?

1) BACK: DEC R0
                JZ BACK
2) BACK: DJNZ RO, BACK

Detailed Solution for Test: Jump, Loop And Call Instructions - Question 10

In the first statement, when the decrements approaches zero then the jump moves back and in second statement, when the result after decrements is not zero, then it jumps back.

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