Computer Architecture Set 4

Computer Science Engineering (CSE)

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This set of Computer Architecture Multiple Choice Questions & Answers (MCQs) focuses on Computer Architecture Set 4

Q1 | Does the Load instruction do the following operation/s?
  • loads the contents of a disc onto a memory location
  • loads the contents of a location onto the accumulators
  • load the contents of the pcb onto the register
  • none of the mentioned
Answer: loads the contents of a location onto the accumulators
Explaination:
the load instruction is basically used to load the contents of a memory location onto a register.
Q2 | Complete the following analogy:- Registers are to RAM’s as Cache’s are to
  • system stacks
  • overlays
  • page table
  • tlb
Answer: tlb
Explaination:
none.
Q3 | The BOOT sector files of the system are stored in                         
  • harddisk
  • rom
  • ram
  • fast solid state chips in the motherboard
Answer: rom
Explaination:
the files which are required for the starting up of a system are stored on the rom.
Q4 | The transfer of large chunks of data with the involvement of the processor is done by                 
  • dma controller
  • arbitrator
  • user system programs
  • none of the mentioned
Answer: dma controller
Explaination:
this mode of transfer involves the transfer of a large block of data from the memory.
Q5 | Which of the following techniques used to effectively utilize main memory?
  • address binding
  • dynamic linking
  • dynamic loading
  • both dynamic linking and loading
Answer: dynamic loading
Explaination:
in this method only when the routine is required is loaded and hence saves memory.
Q6 | RTN stands for                         
  • register transfer notation
  • register transmission notation
  • regular transmission notation
  • regular transfer notation
Answer: register transfer notation
Explaination:
this is the way of writing the assembly language code with the help of register notations.
Q7 | The instruction, Add Loc,R1 in RTN is
  • addsetcc loc+r1
  • r1=loc+r1
  • not possible to write in rtn
  • r1<-[loc]+[r1]
Answer: r1<-[loc]+[r1]
Explaination:
none.
Q8 | Can you perform an addition on three operands simultaneously in ALN using Add instruction?
  • yes
  • not possible using add, we’ve to use addsetcc
  • not permitted
  • none of the mentioned
Answer: not permitted
Explaination:
you cannot perform an addition on three operands simultaneously because the third operand is where the result is stored.
Q9 | The instruction, Add R1,R2,R3 in RTN is                 
  • r3=r1+r2+r3
  • r3<-[r1]+[r2]+[r3]
  • r3=[r1]+[r2]
  • r3<-[r1]+[r2]
Answer: r3<-[r1]+[r2]
Explaination:
in rtn the first operand is the destination and the second operand is the source.
Q10 | In a system, which has 32 registers the register id is                       long.
  • 16 bit
  • 8 bits
  • 5 bits
  • 6 bits
Answer: 5 bits
Explaination:
the id is the name tag given to each of the registers and used to identify them.
Q11 | While using the iterative construct (Branching) in execution                              instruction is used to check the condition.
  • testandset
  • branch
  • testcondn
  • none of the mentioned
Answer: branch
Explaination:
branch instruction is used to check the test condition and to perform the memory jump with the help of offset.
Q12 | The condition flag Z is set to 1 to indicate                 
  • the operation has resulted in an error
  • the operation requires an interrupt call
  • the result is zero
  • there is no empty register available
Answer: the result is zero
Explaination:
this condition flag is used
Q13 |                       converts the programs written in assembly language into machine instructions.
  • machine compiler
  • interpreter
  • assembler
  • converter
Answer: assembler
Explaination:
an assembler is a software used to convert the programs into machine instructions.
Q14 | The instructions like MOV or ADD are called as               
  • op-code
  • operators
  • commands
  • none of the mentioned
Answer: op-code
Explaination:
this op – codes tell the
Q15 | The assembler directive EQU, when used in the instruction: Sum EQU 200 does                   
  • finds the first occurrence of sum and assigns value 200 to it
  • replaces every occurrence of sum with 200
  • re-assigns the address of sum by adding 200 to its original address
  • assigns 200 bytes of memory starting the location of sum
Answer: replaces every occurrence of sum with 200
Explaination:
this basically is used to replace the variable with a constant value.
Q16 | The directive used to perform initialization before the execution of the code is               
  • reserve
  • store
  • dataword
  • equ
Answer: dataword
Explaination:
none.
Q17 |            directive is used to specify and assign the memory required for the block of code.
  • allocate
  • assign
  • set
  • reserve
Answer: reserve
Explaination:
this instruction is used to allocate a block of memory and to store the object code of the program there.
Q18 |            directive specifies the end of execution of a program.
  • end
  • return
  • stop
  • terminate
Answer: return
Explaination:
this instruction directive is used to terminate the program execution.
Q19 | The last statement of the source program should be                 
  • stop
  • return
  • op
  • end
Answer: end
Explaination:
this enables the processor to load some other process.
Q20 | The assembler stores all the names and their corresponding values in               
  • special purpose register
  • symbol table
  • value map set
  • none of the mentioned
Answer: symbol table
Explaination:
the table where the assembler stores the variable names along with their corresponding memory locations and values.
Q21 | The assembler stores the object code in               
  • main memory
  • cache
  • ram
  • magnetic disk
Answer: magnetic disk
Explaination:
after compiling the object code, the assembler stores it in the magnetic disk and waits for further execution.
Q22 | The utility program used to bring the object code into memory for execution is               
  • loader
  • fetcher
  • extractor
  • linker
Answer: loader
Explaination:
the program is used to load the program into memory.
Q23 | To overcome the problems of the assembler in dealing with branching code we use            
  • interpreter
  • debugger
  • op-assembler
  • two-pass assembler
Answer: two-pass assembler
Explaination:
this creates entries into the symbol table first and then creates the object code.
Q24 | The return address of the Sub-routine is pointed to by                 
  • ir
  • pc
  • mar
  • special memory registers
Answer: pc
Explaination:
the return address from the subroutine is pointed to by the pc.
Q25 | The location to return to, from the subroutine is stored in                 
  • tlb
  • pc
  • mar
  • link registers
Answer: link registers
Explaination:
the registers store the return address of the routine and is pointed to by the pc.
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