Load and store instructions (LDR, STR) in ARM Architecture - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to load a value from memory into register R0.

ARM Architecture

LDR R0, [1]

Drag options to blanks, or click blank then click option'

AR0

BR1

C#4

D[R1]

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting to use square brackets around the address register.

Using immediate values instead of memory addresses.

The LDR instruction loads the value from the memory address pointed to by R1 into R0. The correct syntax uses square brackets around the register holding the address.

2fill in blank

medium

Complete the code to store the value from register R2 into the memory address pointed to by R3.

ARM Architecture

STR [1], [R3]

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AR3

BR2

C#8

DR0

Attempts:

3 left

💡 Hint

Common Mistakes

Using the address register as the source operand.

Omitting the square brackets around the address register.

The STR instruction stores the value from register R2 into the memory address pointed to by R3. The first operand is the source register holding the value to store.

3fill in blank

hard

Fix the error in the load instruction to correctly load from the address in R4 with an offset of 12 bytes.

ARM Architecture

LDR R5, [R4[1]]

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A, #4

B#12

C, #12

D#4

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting the comma before the offset.

Placing the offset outside the square brackets.

The correct syntax for loading with an offset is to include a comma and then the immediate offset value inside the brackets. So it should be [R4, #12].

4fill in blank

hard

Fill both blanks to store the value from R7 into the memory address at R8 plus an offset of 20 bytes.

ARM Architecture

STR [1], [[2], #20]

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AR7

BR8

CR9

DR6

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the source and base registers.

Using the wrong register for the base address.

The STR instruction stores the value from R7 into the memory address computed by adding 20 bytes to the address in R8. The syntax requires the source register first, then the base register with offset.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps register names to their load/store instruction types based on whether they use LDR or STR.

ARM Architecture

{ [1]: '[2]' for [3] in ['R0', 'R1', 'R2'] }

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Areg

BLDR

DSTR

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable name in the comprehension.

Confusing the instruction type strings.

This comprehension creates a dictionary where each register name is a key, and the value is the string 'LDR' indicating the load instruction type. The variable 'reg' iterates over the list of registers.

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