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MCQOPTIONS
This section includes 950 Mcqs, each offering curated multiple-choice questions to sharpen your Computer Science Engineering (CSE) knowledge and support exam preparation. Choose a topic below to get started.
| 851. |
The order in which the return addresses are generated and used is |
| A. | lifo |
| B. | fifo |
| C. | random |
| D. | highest priority |
| Answer» B. fifo | |
| 852. |
When parameters are being passed on to the subroutines they are stored in |
| A. | registers |
| B. | memory locations |
| C. | processor stacks |
| D. | all of the mentioned |
| Answer» E. | |
| 853. |
What is subroutine nesting? |
| A. | having multiple subroutines in a program |
| B. | using a linking nest statement to put many subroutines under the same name |
| C. | having one routine call the other |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| 854. |
The location to return to, from the subroutine is stored in |
| A. | tlb |
| B. | pc |
| C. | mar |
| D. | link registers |
| Answer» E. | |
| 855. |
The return address of the Sub-routine is pointed to by |
| A. | ir |
| B. | pc |
| C. | mar |
| D. | special memory registers |
| Answer» C. mar | |
| 856. |
To overcome the problems of the assembler in dealing with branching code we use |
| A. | interpreter |
| B. | debugger |
| C. | op-assembler |
| D. | two-pass assembler |
| Answer» E. | |
| 857. |
The utility program used to bring the object code into memory for execution is |
| A. | loader |
| B. | fetcher |
| C. | extractor |
| D. | linker |
| Answer» B. fetcher | |
| 858. |
The assembler stores the object code in |
| A. | main memory |
| B. | cache |
| C. | ram |
| D. | magnetic disk |
| Answer» E. | |
| 859. |
The last statement of the source program should be |
| A. | stop |
| B. | return |
| C. | op |
| D. | end |
| Answer» E. | |
| 860. |
           directive specifies the end of execution of a program. |
| A. | end |
| B. | return |
| C. | stop |
| D. | terminate |
| Answer» C. stop | |
| 861. |
The assembler stores all the names and their corresponding values in |
| A. | special purpose register |
| B. | symbol table |
| C. | value map set |
| D. | none of the mentioned |
| Answer» C. value map set | |
| 862. |
The assembler directive EQU, when used in the instruction: Sum EQU 200 does |
| A. | finds the first occurrence of sum and assigns value 200 to it |
| B. | replaces every occurrence of sum with 200 |
| C. | re-assigns the address of sum by adding 200 to its original address |
| D. | assigns 200 bytes of memory starting the location of sum |
| Answer» C. re-assigns the address of sum by adding 200 to its original address | |
| 863. |
The directive used to perform initialization before the execution of the code is |
| A. | reserve |
| B. | store |
| C. | dataword |
| D. | equ |
| Answer» D. equ | |
| 864. |
           directive is used to specify and assign the memory required for the block of code. |
| A. | allocate |
| B. | assign |
| C. | set |
| D. | reserve |
| Answer» E. | |
| 865. |
The instructions like MOV or ADD are called as |
| A. | op-code |
| B. | operators |
| C. | commands |
| D. | none of the mentioned |
| Answer» B. operators | |
| 866. |
                      converts the programs written in assembly language into machine instructions. |
| A. | machine compiler |
| B. | interpreter |
| C. | assembler |
| D. | converter |
| Answer» D. converter | |
| 867. |
The condition flag Z is set to 1 to indicate |
| A. | the operation has resulted in an error |
| B. | the operation requires an interrupt call |
| C. | the result is zero |
| D. | there is no empty register available |
| Answer» D. there is no empty register available | |
| 868. |
While using the iterative construct (Branching) in execution                             instruction is used to check the condition. |
| A. | testandset |
| B. | branch |
| C. | testcondn |
| D. | none of the mentioned |
| Answer» C. testcondn | |
| 869. |
In a system, which has 32 registers the register id is                       long. |
| A. | 16 bit |
| B. | 8 bits |
| C. | 5 bits |
| D. | 6 bits |
| Answer» D. 6 bits | |
| 870. |
Can you perform an addition on three operands simultaneously in ALN using Add instruction? |
| A. | yes |
| B. | not possible using add, we’ve to use addsetcc |
| C. | not permitted |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| 871. |
RTN stands for |
| A. | register transfer notation |
| B. | register transmission notation |
| C. | regular transmission notation |
| D. | regular transfer notation |
| Answer» B. register transmission notation | |
| 872. |
The transfer of large chunks of data with the involvement of the processor is done by |
| A. | dma controller |
| B. | arbitrator |
| C. | user system programs |
| D. | none of the mentioned |
| Answer» B. arbitrator | |
| 873. |
Which of the following techniques used to effectively utilize main memory? |
| A. | address binding |
| B. | dynamic linking |
| C. | dynamic loading |
| D. | both dynamic linking and loading |
| Answer» D. both dynamic linking and loading | |
| 874. |
The BOOT sector files of the system are stored in |
| A. | harddisk |
| B. | rom |
| C. | ram |
| D. | fast solid state chips in the motherboard |
| Answer» C. ram | |
| 875. |
Complete the following analogy:- Registers are to RAM’s as Cache’s are to |
| A. | system stacks |
| B. | overlays |
| C. | page table |
| D. | tlb |
| Answer» E. | |
| 876. |
Does the Load instruction do the following operation/s? |
| A. | loads the contents of a disc onto a memory location |
| B. | loads the contents of a location onto the accumulators |
| C. | load the contents of the pcb onto the register |
| D. | none of the mentioned |
| Answer» C. load the contents of the pcb onto the register | |
| 877. |
The unit which acts as an intermediate agent between memory and backing store to reduce process time is |
| A. | tlb’s |
| B. | registers |
| C. | page tables |
| D. | cache |
| Answer» E. | |
| 878. |
The technique used to store programs larger than the memory is |
| A. | overlays |
| B. | extension registers |
| C. | buffers |
| D. | both extension registers and buffers |
| Answer» B. extension registers | |
| 879. |
When generating physical addresses from a logical address the offset is stored in |
| A. | translation look-aside buffer |
| B. | relocation register |
| C. | page table |
| D. | shift register |
| Answer» C. page table | |
| 880. |
If we want to perform memory or arithmetic operations on data in Hexa- decimal mode then we use                     symbol before the operand. |
| A. | ~ |
| B. | ! |
| C. | $ |
| D. | * |
| Answer» D. * | |
| 881. |
Add #%01011101,R1 , when this instruction is executed then |
| A. | the binary addition between the operands takes place |
| B. | the numerical value represented by the binary value is added to the value of r1 |
| C. | the addition doesn’t take place, whereas this is similar to a mov instruction |
| D. | none of the mentioned |
| Answer» B. the numerical value represented by the binary value is added to the value of r1 | |
| 882. |
Physical memory is divided into sets of finite size called as |
| A. | frames |
| B. | pages |
| C. | blocks |
| D. | vectors |
| Answer» B. pages | |
| 883. |
During the transfer of data between the processor and memory we use |
| A. | cache |
| B. | tlb |
| C. | buffers |
| D. | registers |
| Answer» E. | |
| 884. |
To get the physical address from the logical address generated by CPU we use |
| A. | mar |
| B. | mmu |
| C. | overlays |
| D. | tlb |
| Answer» C. overlays | |
| 885. |
           method is used to map logical addresses of variable length onto physical memory. |
| A. | paging |
| B. | overlays |
| C. | segmentation |
| D. | paging with segmentation |
| Answer» D. paging with segmentation | |
| 886. |
When using the Big Endian assignment to store a number, the sign bit of the number is stored in |
| A. | the higher order byte of the word |
| B. | the lower order byte of the word |
| C. | can’t say |
| D. | none of the mentioned |
| Answer» B. the lower order byte of the word | |
| 887. |
The type of memory assignment used in Intel processors is |
| A. | little endian |
| B. | big endian |
| C. | medium endian |
| D. | none of the mentioned |
| Answer» B. big endian | |
| 888. |
If a system is 64 bit machine, then the length of each word will be |
| A. | 4 bytes |
| B. | 8 bytes |
| C. | 16 bytes |
| D. | 12 bytes |
| Answer» C. 16 bytes | |
| 889. |
The collection of the above mentioned entities where data is stored is called |
| A. | block |
| B. | set |
| C. | word |
| D. | byte |
| Answer» D. byte | |
| 890. |
The smallest entity of memory is called |
| A. | cell |
| B. | block |
| C. | instance |
| D. | unit |
| Answer» B. block | |
| 891. |
In a normal n-bit adder, to find out if an overflow as occurred we make use of |
| A. | counter |
| B. | flip flop |
| C. | shift register |
| D. | push down stack |
| Answer» D. push down stack | |
| 892. |
The most efficient method followed by computers to multiply two unsigned numbers is |
| A. | booth algorithm |
| B. | bit pair recording of multipliers |
| C. | restoring algorithm |
| D. | non restoring algorithm |
| Answer» C. restoring algorithm | |
| 893. |
For the addition of large integers, most of the systems make use of |
| A. | fast adders |
| B. | full adders |
| C. | carry look-ahead adders |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| 894. |
In some pipelined systems, a different instruction is used to add to numbers which can affect the flags upon |
| A. | and gate |
| B. | nand gate |
| C. | nor gate |
| D. | xor gate |
| Answer» E. | |
| 895. |
The register used to store the flags is called as |
| A. | flag register |
| B. | status register |
| C. | test register |
| D. | log register |
| Answer» C. test register | |
| 896. |
The processor keeps track of the results of its operations using flags called |
| A. | conditional code flags |
| B. | test output flags |
| C. | type flags |
| D. | none of the mentioned |
| Answer» B. test output flags | |
| 897. |
When we perform subtraction on -7 and 1 the answer in 2’s complement form is |
| A. | 1010 |
| B. | 1110 |
| C. | 0110 |
| D. | 1000 |
| Answer» E. | |
| 898. |
Which method of representation has two representations for ‘0’? |
| A. | sign-magnitude |
| B. | 1’s complement |
| C. | 2’s complement |
| D. | none of the mentioned |
| Answer» B. 1’s complement | |
| 899. |
Which method/s of representation of numbers occupies a large amount of memory than others? |
| A. | sign-magnitude |
| B. | 1’s complement |
| C. | 2’s complement |
| D. | 1’s & 2’s compliment |
| Answer» B. 1’s complement | |
| 900. |
           addressing mode is most suitable to change the normal sequence of execution of instructions. |
| A. | relative |
| B. | indirect |
| C. | index with offset |
| D. | immediate |
| Answer» B. indirect | |