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ARM Architectureknowledge~20 mins

General purpose registers (R0-R12) in ARM Architecture - Practice Problems & Coding Challenges

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Challenge - 5 Problems
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📋 Factual
intermediate
2:00remaining
Purpose of General Purpose Registers
What is the primary role of the general purpose registers R0 to R12 in ARM architecture?
AThey store temporary data and intermediate results during program execution.
BThey hold the program counter and status flags.
CThey are used exclusively for stack management.
DThey store floating-point numbers only.
Attempts:
2 left
💡 Hint

Think about what registers typically do in a CPU during calculations.

🧠 Conceptual
intermediate
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Register Usage in Function Calls
Which registers among R0 to R12 are typically used to pass arguments to functions in ARM calling conventions?
AR10 to R12
BR4 to R7
CR8 to R12
DR0 to R3
Attempts:
2 left
💡 Hint

Consider the first few registers used for passing data to functions.

🔍 Analysis
advanced
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Effect of Using R7 in ARM Assembly
In ARM assembly, what is a common special use of register R7 in many operating systems?
AIt is used to hold the system call number when making system calls.
BIt stores the return address after a function call.
CIt acts as the stack pointer.
DIt is reserved for floating-point operations.
Attempts:
2 left
💡 Hint

Think about how system calls are made in ARM Linux.

❓ Comparison
advanced
2:00remaining
Difference Between R12 and R13
Which statement correctly distinguishes the roles of R12 and R13 in ARM architecture?
AR12 is the program counter, and R13 is a general purpose register.
BR12 is a general purpose register, while R13 is typically used as the stack pointer.
CBoth R12 and R13 are used exclusively for floating-point operations.
DR12 is used for system calls, and R13 holds the link register.
Attempts:
2 left
💡 Hint

Recall the special registers related to stack and program flow.

❓ Reasoning
expert
2:00remaining
Register Preservation Across Function Calls
Which registers among R0 to R12 must a function preserve (save and restore) if it uses them, according to the ARM Procedure Call Standard (AAPCS)?
AR0 to R12
BR0 to R3
CR4 to R11
DR8 to R12 only
Attempts:
2 left
💡 Hint

Think about which registers are caller-saved and which are callee-saved.

Practice

(1/5)
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, R3
MOV R3, R2
MOV R0, R2
B. MOV R0, R2
MOV R2, R3
MOV R3, R0
C. MOV R0, R3
MOV R3, R2
MOV R3, R0
D. MOV R0, R2
MOV R3, R2
MOV 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
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