Addressing modes (immediate, register, offset) in ARM Architecture - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Understanding Immediate Addressing Mode

In ARM assembly, which statement best describes the immediate addressing mode?

A The operand is a constant value encoded directly in the instruction.

B The operand is stored in a memory location addressed by a register.

C The operand is located at an address computed by adding a register and an offset.

D The operand is fetched from a register without any modification.

💡 Hint

Think about where the operand value comes from in immediate mode.

Immediate addressing mode uses a constant value directly in the instruction itself. This means the operand is part of the instruction, not stored in memory or a register.

📋 Factual

intermediate

Register Addressing Mode Usage

Which of the following best describes the register addressing mode in ARM architecture?

A The instruction uses a fixed memory address as the operand.

B The operand is the content of a register specified in the instruction.

C The operand is computed by adding an immediate value to a register.

D The operand is stored in a register offset by another register.

💡 Hint

Consider where the operand is located in register addressing mode.

Register addressing mode means the operand is the value inside a register. The instruction specifies which register to use.

🔍 Analysis

advanced

Offset Addressing Mode Calculation

Given the ARM instruction LDR R1, [R2, #4], what address is used to load the value into R1?

A The address stored in R2 plus 4 bytes.

B The address stored in R1 plus 4 bytes.

C The address stored in R2 minus 4 bytes.

D The immediate value 4 is loaded directly into R1.

💡 Hint

Look at the base register and the offset in the instruction.

The instruction loads data from the memory address calculated by adding 4 to the value in register R2. This is offset addressing mode.

❓ Comparison

advanced

Comparing Addressing Modes in ARM

Which addressing mode allows the operand to be a memory address calculated by adding two registers together?

A Register offset addressing mode

B Register indirect addressing mode

C Indexed addressing mode

D Immediate addressing mode

💡 Hint

Think about which mode uses two registers to compute the address.

Indexed addressing mode computes the memory address by adding two registers, often a base and an index register.

❓ Reasoning

expert

Determining the Result of Mixed Addressing Modes

Consider the ARM instruction STR R0, [R1, R2, LSL #2]. What does this instruction do?

A Stores the value of R0 into the memory address at R2 shifted left by 2 bits.

B Stores the value of R0 into the memory address at R1 plus 2.

C Loads the value from the memory address at R1 plus R2 into R0.

D Stores the value of R0 into the memory address at R1 plus R2 shifted left by 2 bits.

💡 Hint

Focus on how the address is calculated using the shift operation.

This instruction stores the value in R0 into the memory address computed by adding R1 and the value of R2 shifted left by 2 bits (multiplied by 4). This is a form of register offset addressing with scaling.

Practice

1. Which addressing mode uses a fixed number directly inside the instruction to provide data?

easy

A. Immediate addressing mode

B. Register addressing mode

C. Offset addressing mode

D. Indirect addressing mode

Solution

1. Step 1: Understand immediate addressing mode

   Immediate mode means the instruction contains the actual number (constant) to use.

2. Step 2: Compare with other modes

   Register mode uses CPU registers, offset mode calculates address by adding offset to a register, indirect mode uses memory address stored in a register.

3. Final Answer:

   Immediate addressing mode -> Option A

4. Quick Check:

   Fixed number inside instruction = Immediate mode [OK]

Hint: Immediate mode has the number right inside the instruction [OK]

Common Mistakes:

• Confusing immediate with register mode
• Thinking offset mode uses fixed numbers directly
• Mixing indirect addressing with immediate

2. Which of the following is the correct syntax for using register addressing mode in ARM assembly to move data from register R1 to R0?

easy

A. MOV R0, [R1]

B. MOV R0, #R1

C. MOV R0, R1

D. MOV [R0], R1

Solution

1. Step 1: Understand register addressing syntax

   Register addressing uses register names directly without # or brackets.

2. Step 2: Analyze options

   MOV R0, R1 uses MOV R0, R1 which copies data from R1 to R0 correctly. MOV R0, #R1 wrongly uses # which means immediate. MOV R0, [R1] uses brackets which means memory access. MOV [R0], R1 tries to move to memory location incorrectly.

3. Final Answer:

   MOV R0, R1 -> Option C

4. Quick Check:

   Register names without # or [] = Register mode [OK]

Hint: Register mode uses register names directly, no # or brackets [OK]

Common Mistakes:

• Using # before register name
• Using brackets [] for register mode
• Confusing immediate and register syntax

3. Given the ARM instruction LDR R2, [R3, #4], what does this instruction do?

medium

A. Loads the immediate value 4 into R2

B. Loads the value from memory address in R3 plus 4 into R2

C. Moves the value from R3 to R2

D. Stores the value of R2 into memory address R3 plus 4

Solution

1. Step 1: Identify addressing mode in LDR instruction

   The syntax [R3, #4] means offset addressing mode: base register R3 plus immediate offset 4.

2. Step 2: Understand LDR operation

   LDR loads data from memory address calculated by R3 + 4 into register R2.

3. Final Answer:

   Loads the value from memory address in R3 plus 4 into R2 -> Option B

4. Quick Check:

   Offset mode loads from (R3 + 4) into R2 [OK]

Hint: Brackets with # offset means load from memory address plus offset [OK]

Common Mistakes:

• Thinking #4 is immediate load into R2
• Confusing load with store instruction
• Ignoring the offset in brackets

4. Identify the error in this ARM instruction using addressing modes: ADD R1, R2, #R3

medium

A. Incorrect use of # before register R3

B. Missing brackets around R3

C. Wrong destination register

D. Immediate value cannot be used with ADD

Solution

1. Step 1: Analyze the use of # in ARM instructions

   The # symbol is used only before immediate (constant) values, not registers.

2. Step 2: Check the instruction syntax

   Here #R3 is invalid because R3 is a register, not an immediate value. Correct syntax is ADD R1, R2, R3 without #.

3. Final Answer:

   Incorrect use of # before register R3 -> Option A

4. Quick Check:

   # only for constants, not registers [OK]

Hint: # is only for numbers, never before register names [OK]

Common Mistakes:

• Using # before register names
• Adding brackets unnecessarily
• Confusing immediate and register operands

5. You want to load a value from memory address calculated by adding 8 to the value in register R4, then add an immediate value 5 to it and store in R0. Which sequence of ARM instructions correctly uses addressing modes?

hard

A. LDR R1, [R4, R5]
ADD R0, R1, #8

B. LDR R0, [R4, #5]
ADD R0, R0, #8

C. LDR R0, [R4]
ADD R0, R0, R5

D. LDR R1, [R4, #8]
ADD R0, R1, #5

Solution

1. Step 1: Load from memory address R4 + 8

   Use offset addressing mode: LDR R1, [R4, #8] loads from address R4 + 8 into R1.

2. Step 2: Add immediate 5 to loaded value

   ADD R0, R1, #5 adds immediate 5 to R1 and stores result in R0.

3. Final Answer:

   LDR R1, [R4, #8] ADD R0, R1, #5 -> Option D

4. Quick Check:

   Offset load then immediate add = LDR R1, [R4, #8] ADD R0, R1, #5 [OK]

Hint: Load with offset first, then add immediate value [OK]

Common Mistakes:

• Mixing immediate and offset values incorrectly
• Using register instead of immediate for offset
• Loading directly into R0 then adding wrong offset