Compilers

Runtime Storage Location

Question 1
Marks : +2 | -2
Pass Ratio : 100%
Translation from symbolic program into Binary is done in __________
Two passes
Directly
Three passes
Four passes
Explanation:
Two passes are required.
Question 2
Marks : +2 | -2
Pass Ratio : 100%
The time interval between adjacent bits is called the __________
Word-time
Bit-time
Turnaround time
Slice time
Explanation:
Bit time is the interval time here.
Question 3
Marks : +2 | -2
Pass Ratio : 100%
Memory unit accessed by content is called ______
Read only memory
Programmable Memory
Virtual Memory
Associative Memory
Explanation:
Associative Emory is accessed by content.
Question 4
Marks : +2 | -2
Pass Ratio : 100%
A group of bits that tell the computer to perform a specific operation is known as ________
Instruction code
Micro-operation
Accumulator
Register
Explanation:
Instruction code is the set of specific tasks to be performed.
Question 5
Marks : +2 | -2
Pass Ratio : 100%
The average time required to reach a storage location in memory and obtain its contents is called __________
Latency time
Access time
Turnaround time
Response time
Explanation:
Access time is the time from the start of one storage device access to the time when the next access can be started.
Question 6
Marks : +2 | -2
Pass Ratio : 100%
The load instruction is mostly used to designate a transfer from memory to a processor register known as _______
Accumulator
Instruction Register
Program counter
Memory address Register
Explanation:
Accumulator is the process register.
Question 7
Marks : +2 | -2
Pass Ratio : 100%
‘Aging registers’ are _______
Counters which indicate how long ago their associated pages have been Referenced
Registers which keep track of when the program was last accessed
Counters to keep track of last accessed instruction
Counters to keep track of the latest data structures referred
Explanation:
This is known as aging.
Question 8
Marks : +2 | -2
Pass Ratio : 100%
The multiplicand register & multiplier register of a hardware circuit implementing booth’s algorithm have (11101) & (1100). The result shall be ______
(812)10
(-12)10
(12)10
(-812)10
Explanation:
On multiplying we get 81210.
Question 9
Marks : +2 | -2
Pass Ratio : 100%
The instruction ‘ORG O’ is a ______
Machine Instruction
Pseudo instruction
High level instruction
Memory instruction
Explanation:
It is a pseudo instruction.
Question 10
Marks : +2 | -2
Pass Ratio : 100%
The circuit converting binary data in to decimal is_____
Encoder
Multiplexer
Decoder
Code converter
Explanation:
It is the code converter.