Machine Instructions and Addressing Modes MCQ Quiz in मल्याळम - Objective Question with Answer for Machine Instructions and Addressing Modes - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is "option 3".

CONCEPT:

To obtain operand value from memory, the address field of an instruction is used by the CPU.

In single address instruction, one of the operands is stored in the accumulator & the other operand may be either in register or memory. 

EXPLANATION:

All the operations will be performed in the Accumulator register(AC).

The load operation is used to fetch the value from register or memory to accumulator.

The store operation is used to store the value from the accumulator to register or memory.

The one address instructions for the given equations are:​

Hence, the correct answer is "option 3".

In assembly language mnemonics are used to represent operation codes.

Opcodes (operation codes) are represented by abbreviations, called mnemonics that indicate the operation.

Mnemonic codes:

• Mnemonic codes are the codes that can be remembered comparatively easily and that aids its user in recalling the information it represents.
• Mnemonics codes are widely used in computer programming and communication system operations to specify instructions.

Examples:

ADD       Add

SUB        Subtract

MUL      Multiply

DIV         Divide

LOAD     Load data from memory

STOR     Store data to memory

Assembly language: It is a low-level programming language that uses symbols, variables, and functions that work directly with CPU.

High-level language: It is a human-friendly language that uses variables and functions, independent of the computer architecture.

Data:

Number of registers = 28

Number of instructions = 35

Formula:

Number of bits =⌈ log₂N ⌉

Calculation:

Number of bits to address register = ⌈ log₂ 28 ⌉ = 5

Opcode size = ⌈ log₂ 35 ⌉ = 6 bits

source register = destination register = 5 bits

Total bits per instruction = 6 + 5 + 5 + 16 = 32 bits

In terms of bytes= \(\frac{{32}}{8}\)= 4 bytes

Total instructions = 2000

Instruction set: ADD, SUB, LDA, STA, CLR

There are 5 opcodes possible at any time within an instruction. Hence each opcode needs to denote with a unique binary sequence.

Consider the below table:

Therefore minimum 3 bits is needed for opcode.

Shortcuts:

Number of opcodes = n

Data:

Instruction length = 20 bits

Total address possible = 2²⁰

Operand specifiers = 6 bits

0 – address = 0

1 – address = y × 2⁶ 

2 – address = K × 26 × 26= 212 K

Formula:

0 – address  + 1 – address + 2 – address ≤ Total Address

Calculation:

0 + y× 2⁶  + 212 K ≤ 220

y ≤ 2¹⁴ – 2⁶ K 

y_max = 214 – 26 K 

∴ option 1 is correct

The correct answer is option 4:

Key Points

In stack organized architecture push and pop instruction is needs a address field to specify the location of data for pushing into the stack and destination location during pop operation but for logic and arithmetic operation the instruction does not need any address field as it operates on the top two data available in the stack.

Data:

Instruction length = 16 bits

Number of registers = 64

Bits to represent register = ⌈ log₂ (64) ⌉ = 6

Explanation

I-type instruction format:

R-type instruction format =

Instruction length is given 16, therefore maximum possible encodings = 2¹⁶

Number of I-type opcodes = 8

I-type Instructions encoding + R-type instructions encoding = Total instructions

Assume the number of R-type opcodes = x

Therefore,

(8 × 2⁶ × 2⁴) + (x × 2⁶ × 2⁶) = 2¹⁶

÷ b 2¹² on both side

∴ 2 + x = 2⁴

∴ x = 14.

The correct answer is option 1, option 3 and option 4.

Concept:

Option 1: In the immediate addressing mode the operand is placed in the instruction itself.

True, An immediate mode instruction has an operand field rather than an address field. The operand field contains the actual operand to be used in conjunction with the operation specified in the instruction.

Option 2: One-byte machine instruction consists of the only operand.

False, The machine instructions which consist of the only opcode are called one-byte machine instructions.

Option 3: Indirect addressing mode is suitable for implementing pointers in C.

True, Indirect addressing mode is suitable for implementing pointers in C. The instruction includes the address of the place where the target address is stored in this style of addressing. As a result, it is indirectly storing the target site's address in another memory location.

Option 4: Displacement addressing mode is similar to the register indirect addressing mode.

True, The displacement addressing mode is equal to the register indirect addressing mode, with the exception that the effective address of the operand is formed by adding an offset (or displacement) maintained in the instruction to the contents of the register provided in the instruction.

Hence the correct answer is option 1, option 3 and option 4.

CPU will pick a program from the ready queue maintained by Operating System and will start executing it. At that time it is said to be in system mode because it is executing the normal program for the system.

In case if interrupt comes in between then after executing the current process(high priority) CPU will search for the interrupt to execute.
The interrupt is a signal emitted by hardware or software when a process or an event needs immediate attention.

Let's analyze the options one by one-
Option 1-When CPU will service the interrupt request at that time it is said to be in interrupt mode.

Option 2- This is the correct answer as explained above.

Option 3- There is nothing called "Half mode " in operating system execution.

Option 4- Simplex mode is a concept in Computer Network where the communication between sender and receiver occur only in one direction.

The correct answer is option 2.

Concept:

The given data,

Physical Address = 64 bit and it is byte-addressable

1 Word = 64 bit = 8 bytes

2 word = 128 bit = 16 bytes

Base location = 2000

If an interrupt occurs the CPU has been halted after execution of the HALT instruction, the return address 2040 is saved in the stack.

Hence the correct answer is 2040.