Load and store instructions (LDR, STR) in ARM Architecture - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Understanding the purpose of LDR and STR instructions

What is the primary function of the LDR and STR instructions in ARM architecture?

A Both LDR and STR are used to perform arithmetic operations on registers.

B LDR loads data from memory into a register; STR stores data from a register into memory.

C LDR and STR are used to jump to different parts of the program.

D LDR stores data from a register into memory; STR loads data from memory into a register.

💡 Hint

Think about how data moves between memory and registers.

LDR (Load Register) copies data from memory into a register. STR (Store Register) copies data from a register into memory. They are fundamental for data transfer.

📋 Factual

intermediate

Addressing modes used by LDR and STR

Which addressing mode is NOT commonly used with the LDR and STR instructions in ARM?

A Immediate offset addressing

B Register indirect addressing with optional offset

C Direct register-to-register addressing without memory access

D Indexed addressing with pre- or post-increment

💡 Hint

Consider how LDR and STR interact with memory addresses.

LDR and STR always involve memory access. Direct register-to-register addressing without memory access is not an addressing mode for these instructions.

🔍 Analysis

advanced

Predicting the effect of an LDR instruction

Given the ARM instruction LDR R1, [R2, #4], what happens when this instruction executes?

A The value at the memory address in R2 is loaded into R1, then 4 is added to R2.

B The value in R1 is stored into the memory address in R2 plus 4.

C R1 is loaded with the value 4, and R2 remains unchanged.

D The value at the memory address in R2 plus 4 is loaded into R1.

💡 Hint

Focus on the syntax: [R2, #4] means base register plus offset.

The instruction loads into R1 the value stored at the memory address calculated by adding 4 to the value in R2. It does not modify R2.

❓ Comparison

advanced

Difference between STR and STRB instructions

What is the key difference between the STR and STRB instructions in ARM?

A STR stores a full word (4 bytes), STRB stores a single byte.

B STR stores data in big-endian format, STRB in little-endian.

C STR stores data only to registers, STRB stores data to memory.

D STR is used for loading data, STRB is used for storing data.

💡 Hint

Consider the size of data each instruction handles.

STR stores a 32-bit word from a register into memory. STRB stores only the least significant byte (8 bits) from a register into memory.

❓ Reasoning

expert

Effect of post-increment addressing in STR instruction

Consider the instruction STR R0, [R1], #4. What is the state of register R1 after this instruction executes?

A R1 is increased by 4 after storing R0's value to the original address in R1.

B R1 remains unchanged; the value of R0 is stored at the address in R1 plus 4.

C R1 is increased by 4 before storing R0's value to the new address in R1.

D R1 is decreased by 4 after storing R0's value to the original address in R1.

💡 Hint

Post-increment means the register is updated after the memory operation.

The instruction stores the value of R0 at the address currently in R1, then increments R1 by 4 after the store.

Practice

1. What does the LDR instruction do in ARM architecture?

easy

A. Stores data from a register into memory

B. Loads data from memory into a register

C. Performs arithmetic operations on registers

D. Jumps to a different instruction address

Solution

1. Step 1: Understand the purpose of LDR

   The LDR instruction is used to read or load data from a memory address into a CPU register.

2. Step 2: Differentiate from STR

   STR is the opposite; it stores data from a register back into memory.

3. Final Answer:

   Loads data from memory into a register -> Option B

4. Quick Check:

   LDR = Load from memory [OK]

Hint: LDR loads from memory, STR stores to memory [OK]

Common Mistakes:

• Confusing LDR with STR
• Thinking LDR performs arithmetic
• Assuming LDR changes program flow

2. Which of the following is the correct syntax to store the value of register R1 into the memory address pointed by R0?

easy

A. STR R0, [R1]

B. LDR R1, [R0]

C. STR R1, [R0]

D. LDR R0, [R1]

Solution

1. Step 1: Identify the store instruction

   STR is used to store data from a register into memory.

2. Step 2: Match registers to syntax

   STR R1, [R0] means store the value in R1 into the memory address held in R0.

3. Final Answer:

   STR R1, [R0] -> Option C

4. Quick Check:

   STR stores register to memory [OK]

Hint: STR stores register to memory address in brackets [OK]

Common Mistakes:

• Swapping register and address in syntax
• Using LDR instead of STR for storing
• Confusing source and destination registers

3. Given the following ARM instructions:

STR R2, [R3]
LDR R4, [R3]

What will be the value in register R4 after execution if R2 = 10 and the memory at address in R3 initially contains 5?

medium

A. 10

B. 5

C. Undefined

D. 0

Solution

1. Step 1: Execute STR instruction

   STR R2, [R3] stores the value 10 (from R2) into the memory address pointed by R3, replacing the initial 5.

2. Step 2: Execute LDR instruction

   LDR R4, [R3] loads the updated memory value (now 10) into R4.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Store then load updates R4 to 10 [OK]

Hint: Store first, then load from same address updates register [OK]

Common Mistakes:

• Assuming memory value stays 5 after STR
• Confusing source and destination registers
• Ignoring instruction order

4. Identify the error in the following ARM assembly snippet:

LDR R1, R2
STR R3, [R4]

medium

A. LDR syntax is incorrect; missing brackets around R2

B. STR syntax is incorrect; missing brackets around R3

C. Both instructions are correct

D. LDR should use STR instead

Solution

1. Step 1: Check LDR syntax

   LDR loads from memory address, so the source must be in brackets like [R2]. Here, R2 is missing brackets, so syntax is wrong.

2. Step 2: Check STR syntax

   STR R3, [R4] is correct because it stores R3's value into memory at address in R4.

3. Final Answer:

   LDR syntax is incorrect; missing brackets around R2 -> Option A

4. Quick Check:

   LDR needs brackets for memory address [OK]

Hint: LDR and STR need brackets for memory addresses [OK]

Common Mistakes:

• Omitting brackets for memory operands
• Confusing register direct with memory access
• Thinking STR needs brackets on source register

5. You want to copy a block of 4 words from memory starting at address in R5 to another location starting at address in R6 using ARM instructions. Which sequence correctly uses LDR and STR to achieve this?

hard

A. Use LDR R0, [R5, #4]; STR R0, [R6, #4] once

B. Use LDR R0, [R5]; STR R0, [R6] once only

C. Use STR R0, [R5]; LDR R0, [R6] repeated 4 times

D. Use a loop: LDR R0, [R5], #4; STR R0, [R6], #4 repeated 4 times

Solution

1. Step 1: Understand block copy requirement

   Copying 4 words means loading each word from source and storing it to destination, advancing addresses each time.

2. Step 2: Analyze Use a loop: LDR R0, [R5], #4; STR R0, [R6], #4 repeated 4 times

   Use a loop: LDR R0, [R5], #4; STR R0, [R6], #4 repeated 4 times uses LDR and STR with post-increment addressing ([R5], #4 and [R6], #4) inside a loop repeated 4 times, correctly copying each word and moving pointers.

3. Step 3: Analyze other options

   Use LDR R0, [R5]; STR R0, [R6] once only copies only one word. Use STR R0, [R5]; LDR R0, [R6] repeated 4 times reverses load/store roles. Use LDR R0, [R5, #4]; STR R0, [R6, #4] once copies only one word with offset but no loop.

4. Final Answer:

   Use a loop: LDR R0, [R5], #4; STR R0, [R6], #4 repeated 4 times -> Option D

5. Quick Check:

   Loop with LDR/STR and address increment copies block [OK]

Hint: Use loop with LDR/STR and post-increment for block copy [OK]

Common Mistakes:

• Copying only one word instead of all 4
• Mixing load and store instructions
• Not incrementing addresses after each word