General purpose registers (R0-R12) in ARM Architecture - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the correct general purpose register used for the first argument in ARM calling conventions.

ARM Architecture

MOV [1], #5  ; Move value 5 into the first argument register

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

AR0

BR13

CR7

DR12

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing R13 which is the stack pointer.

Choosing R7 which is sometimes used as frame pointer.

In ARM architecture, R0 is used to pass the first argument to functions.

2fill in blank

medium

Complete the code to store the value of register R5 into memory address pointed by R1.

ARM Architecture

STR [1], [R1]

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

AR0

BR5

CR12

DR7

Attempts:

3 left

💡 Hint

Common Mistakes

Using R0 which might not hold the intended value.

Using R12 which is often used as an intra-procedure-call scratch register.

The STR instruction stores the value from the specified register into the memory address pointed to by the second operand. Here, R5 is the source register.

3fill in blank

hard

Fix the error in the code by selecting the correct register to use as the frame pointer in ARM.

ARM Architecture

MOV [1], SP  ; Set frame pointer to stack pointer

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

AR12

BR0

CR13

DR7

Attempts:

3 left

💡 Hint

Common Mistakes

Using R13 which is the stack pointer.

Using R0 which is for arguments.

In ARM, R7 is commonly used as the frame pointer (FP) in many calling conventions.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps register names to their numbers for R0 to R3.

ARM Architecture

{'R'+str(i): i[1] i in range(4) if i [2] 2}

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

A for

B>

C<

D if

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'if' instead of 'for' in the first blank.

Using '<' instead of '>' in the second blank.

The comprehension iterates over i in range(4) using 'for' and filters with '>' to include only registers with numbers greater than 2.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension mapping registers R8 to R10 to their hexadecimal values if the value is less than 0xA.

ARM Architecture

{'R'+str([1]): hex([2]) for [3] in range(8, 11) if [2] < 0xA}

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

Ai

Cx

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names causing errors.

Using variables not defined in the comprehension.

The variable 'i' is used for iteration and as the key and value in the comprehension. The condition filters values less than 0xA.

Practice

1. Which of the following registers are considered general purpose registers in ARM architecture?

easy

A. R0 to R12

B. R13 to R15

C. Program Counter and Status Register

D. Floating Point Registers

Solution

1. Step 1: Identify general purpose registers in ARM

   Registers R0 through R12 are used for general data storage and manipulation.

2. Step 2: Differentiate special registers

   Registers R13 to R15 and others like Program Counter are special purpose, not general.

3. Final Answer:

   R0 to R12 -> Option A

4. Quick Check:

   General purpose registers = R0-R12 [OK]

Hint: General purpose registers are R0 to R12 only [OK]

Common Mistakes:

• Confusing R13-R15 as general purpose
• Including special registers like Program Counter
• Thinking floating point registers are general purpose

2. Which of the following is the correct way to refer to the 5th general purpose register in ARM assembly?

easy

A. R6

B. R5

C. R4

D. R7

Solution

1. Step 1: Understand register numbering

   Registers start at R0, so the 1st is R0, 2nd is R1, and so on.

2. Step 2: Count to the 5th register

   Counting from R0: R0(1), R1(2), R2(3), R3(4), R4(5). So the 5th register is R4.

3. Step 3: Verify options

   R5 is R5, which is actually the 6th register, so the correct is R4 (R4).

4. Final Answer:

   R4 -> Option C

5. Quick Check:

   5th register = R4 [OK]

Hint: Count from R0 to find the register number [OK]

Common Mistakes:

• Starting count from R1 instead of R0
• Mixing register numbers with their positions
• Choosing R5 thinking it is the 5th register

3. Given the ARM assembly instructions:
MOV R0, #10
ADD R1, R0, #5
SUB R2, R1, R0
What is the final value stored in register R2?

medium

A. 5

B. 15

C. 10

D. 0

Solution

1. Step 1: Execute MOV and ADD instructions

   MOV R0, #10 sets R0 = 10. ADD R1, R0, #5 sets R1 = 10 + 5 = 15.

2. Step 2: Execute SUB instruction

   SUB R2, R1, R0 sets R2 = 15 - 10 = 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   R2 = R1 - R0 = 15 - 10 = 5 [OK]

Hint: Calculate step-by-step register values [OK]

Common Mistakes:

• Mixing up source and destination registers
• Forgetting immediate values in ADD
• Assuming R2 equals R1 or R0 directly

4. Identify the error in the following ARM assembly snippet:
MOV R13, #20
ADD R0, R13, #5
What is the issue?

medium

A. ADD instruction requires three registers, not two registers and an immediate

B. Immediate value #5 is invalid

C. MOV instruction cannot move immediate values

D. R13 is not a general purpose register and should not be used like this

Solution

1. Step 1: Check register usage

   R13 is the stack pointer, a special purpose register, not general purpose.

2. Step 2: Validate instruction correctness

   Using R13 like a general purpose register is incorrect; MOV and ADD with immediate values are valid.

3. Final Answer:

   R13 is not a general purpose register and should not be used like this -> Option D

4. Quick Check:

   R13 is special purpose, not general [OK]

Hint: Remember R13-R15 are special, not general registers [OK]

Common Mistakes:

• Assuming R13 is general purpose
• Thinking immediate values are invalid
• Believing ADD can't use immediate values

5. You want to swap the values of registers R2 and R3 using only R0 as a temporary register. Which sequence of instructions correctly achieves this?

hard

A. MOV R2, R3MOV R3, R2MOV R0, R2

B. MOV R0, R2MOV R2, R3MOV R3, R0

C. MOV R0, R3MOV R3, R2MOV R3, R0

D. MOV R0, R2MOV R3, R2MOV R2, R0

Solution

1. Step 1: Save R2 value temporarily

   MOV R0, R2 stores R2's value in R0 to keep it safe.

2. Step 2: Move R3 value to R2

   MOV R2, R3 copies R3's value into R2, overwriting original R2.

3. Step 3: Restore original R2 value to R3

   MOV R3, R0 moves saved R2 value from R0 into R3, completing the swap.

4. Final Answer:

   MOV R0, R2; MOV R2, R3; MOV R3, R0 -> Option B

5. Quick Check:

   Use temp register R0 to swap R2 and R3 [OK]

Hint: Use R0 as temp: save R2, copy R3 to R2, restore R2 to R3 [OK]

Common Mistakes:

• Overwriting R2 before saving it
• Swapping without a temporary register
• Mixing source and destination registers