Addressing Modes | Practice Problems | COA

Addressing Modes in Computer Architecture-

 

Before you go through this article, make sure that you have gone through the previous article on Addressing Modes.

 

We have discussed-

• Addressing modes are the different ways of specifying the location of an operand.
• Various kinds of addressing modes and their applications.

 

 

Also Read- Syntax Of Addressing Modes

 

In this article, we will discuss practice problems based on addressing modes.

 

PRACTICE PROBLEMS BASED ON ADDRESSING MODES-

 

Problem-01:

 

The most appropriate matching for the following pairs is-

 

Column-1:

X: Indirect addressing

Y: Immediate addressing

Z: Auto decrement addressing

 

Column-2:

1. Loops

2. Pointers

3. Constants

 

1. X-3, Y-2, Z-1
2. X-1, Y-3, Z-2
3. X-2, Y-3, Z-1
4. X-3, Y-1, Z-2

 

Solution-

 

Option (C) is correct.

 

Problem-02:

 

In the absolute addressing mode,

1. The operand is inside the instruction
2. The address of the operand is inside the instruction
3. The register containing the address of the operand is specified inside the instruction
4. The location of the operand is implicit

 

Solution-

 

Option (B) is correct.

 

Problem-03:

 

Which of the following addressing modes are suitable for program relocation at run time?

1. Absolute addressing
2. Base addressing
3. Relative addressing
4. Indirect addressing

 

1. 1 and 4
2. 1 and 2
3. 2 and 3
4. 1, 2 and 4

 

Solution-

 

Option (C) is correct.

 

Problem-04:

 

What is the most appropriate match for the items in the first column with the items in the second column-

 

Column-1:

X: Indirect addressing

Y: Indexed addressing

Z: Base register addressing

 

Column-2:

1. Array implementation

2. Writing relocatable code

3. Passing array as parameter

 

1. X-3, Y-1, Z-2
2. X-2, Y-3, Z-1
3. X-3, Y-2, Z-1
4. X-1, Y-3, Z-2

 

Solution-

 

Option (A) is correct.

 

Problem-05:

 

Which of the following addressing modes permits relocation without any change whatsoever in the code?

1. Indirect addressing
2. Indexed addressing
3. Base register addressing
4. PC relative addressing

 

Solution-

 

Option (C) is correct.

 

Problem-06:

 

Consider a three word machine instruction-

ADD A[R₀], @B

The first operand (destination) “A[R₀]” uses indexed addressing mode with R₀ as the index register. The second operand operand (source) “@B” uses indirect addressing mode. A and B are memory addresses residing at the second and the third words, respectively. The first word of the instruction specifies the opcode, the index register designation and the source and destination addressing modes. During execution of ADD instruction, the two operands are added and stored in the destination (first operand).

The number of memory cycles needed during the execution cycle of the instruction is-

1. 3
2. 4
3. 5
4. 6

 

Solution-

 

For the first operand,

• It uses indexed addressing mode.
• Thus, one memory cycle will be needed to fetch the operand.

 

For the second operand,

• It uses indirect addressing mode.
• Thus, two memory cycles will be needed to fetch the operand.

 

After fetching the two operands,

• The operands will be added and result is stored back in the memory.
• Thus, one memory cycle will be needed to store the result.

 

Total number of memory cycles needed

=  1 + 2 + 1

= 4

 

Thus, Option (B) is correct.

To watch video solution, click here.

 

Problem-07:

 

Consider a hypothetical processor with an instruction of type LW R₁, 20(R₂), which during execution reads a 32-bit word from memory and stores it in a 32-bit register R₁. The effective address of the memory location is obtained by the addition of a constant 20 and the contents of register R₂. Which of the following best reflects the addressing mode implemented by this instruction for operand in memory?

1. Immediate Addressing
2. Register Addressing
3. Register Indirect Scaled Addressing
4. Base Indexed Addressing

 

Solution-

 

Clearly, the instruction uses base indexed addressing mode.

Thus, Option (D) is correct.

 

Problem-08:

 

The memory locations 1000, 1001 and 1020 have data values 18, 1 and 16 respectively before the following program is executed.

 

MOVI	Rs, 1	Move immediate
LOAD	Rd, 1000(Rs)	Load from memory
ADDI	Rd, 1000	Add immediate
STOREI	0(Rd), 20	Store immediate

 

Which of the statements below is TRUE after the program is executed?

1. Memory location 1000 has value 20
2. Memory location 1020 has value 20
3. Memory location 1021 has value 20
4. Memory location 1001 has value 20

 

Solution-

 

Before the execution of program, the memory is-

 

 

Now, let us execute the program instructions one by one-

 

Instruction-01: MOVI R_s, 1

 

• This instruction uses immediate addressing mode.
• The instruction is interpreted as Rs ← 1.
• Thus, value = 1 is moved to the register R_s.

 

Instruction-02: LOAD R_d, 1000(R_s)

 

• This instruction uses displacement addressing mode.
• The instruction is interpreted as R_d ← [1000 + [R_s]].
• Value of the operand = [1000 + [R_s]] = [1000 + 1] = [1001] = 1.
• Thus, value = 1 is moved to the register R_d.

 

Instruction-03: ADDI R_d, 1000

 

• This instruction uses immediate addressing mode.
• The instruction is interpreted as R_d ← [R_d] + 1000.
• Value of the operand = [R_d] + 1000 = 1 + 1000 = 1001.
• Thus, value = 1001 is moved to the register R_d.

 

Instruction-04: STOREI 0(R_d), 20

 

• This instruction uses displacement addressing mode.
• The instruction is interpreted as 0 + [R_d] ← 20.
• Value of the destination address = 0 + [R_d] = 0 + 1001 = 1001.
• Thus, value = 20 is moved to the memory location 1001.

 

Thus,

• After the program execution is completed, memory location 1001 has value 20.
• Option (D) is correct.

 

To watch video solution, click here.

 

Problem-09:

 

Consider the following memory values and a one-address machine with an accumulator, what values do the following instructions load into accumulator?

• Word 20 contains 40
• Word 30 contains 50
• Word 40 contains 60
• Word 50 contains 70

Instructions are-

1. Load immediate 20
2. Load direct 20
3. Load indirect 20
4. Load immediate 30
5. Load direct 30
6. Load indirect 30

 

Solution-

 

Instruction-01: Load immediate 20

 

• This instruction uses immediate addressing mode.
• The instruction is interpreted as Accumulator ← 20.
• Thus, value 20 is loaded into the accumulator.

 

Instruction-02: Load direct 20

 

• This instruction uses direct addressing mode.
• The instruction is interpreted as Accumulator ← [20].
• It is given that word 20 contains 40.
• Thus, value 40 is loaded into the accumulator

 

Instruction-03: Load indirect 20

 

• This instruction uses indirect addressing mode.
• The instruction is interpreted as Accumulator ← [[20]].
• It is given that word 20 contains 40 and word 40 contains 60.
• Thus, value 60 is loaded into the accumulator.

 

Instruction-04: Load immediate 30

 

• This instruction uses immediate addressing mode.
• The instruction is interpreted as Accumulator ← 30.
• Thus, value 30 is loaded into the accumulator.

 

Instruction-02: Load direct 30

 

• This instruction uses direct addressing mode.
• The instruction is interpreted as Accumulator ← [30].
• It is given that word 30 contains 50.
• Thus, value 50 is loaded into the accumulator

 

Instruction-03: Load indirect 30

 

• This instruction uses indirect addressing mode.
• The instruction is interpreted as Accumulator ← [[30]].
• It is given that word 30 contains 50 and word 50 contains 70.
• Thus, value 70 is loaded into the accumulator.

 

To watch video solution, click here.

 

Next Article- System Bus in Computer Architecture

 

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