Bird
Raised Fist0
ARM Architectureknowledge~5 mins

General purpose registers (R0-R12) in ARM Architecture

Choose your learning style10 modes available
Introduction

General purpose registers hold data and addresses that the processor uses to perform tasks quickly. They help the CPU work efficiently by storing temporary information.

When performing arithmetic calculations like addition or subtraction.
When temporarily storing data during program execution.
When passing arguments to functions or receiving results.
When manipulating memory addresses for data access.
When saving intermediate results in complex operations.
Core Concept
ARM Architecture
R0, R1, R2, ..., R12

Registers R0 to R12 are used for general tasks and can hold any data the program needs.

They are 32-bit wide in ARM architecture, meaning they can store 32-bit numbers.

Key Points
This moves the number 5 into register R0.
ARM Architecture
MOV R0, #5
This adds the values in R0 and R2, then stores the result in R1.
ARM Architecture
ADD R1, R0, R2
This stores the value from R3 into the memory address pointed to by R4.
ARM Architecture
STR R3, [R4]
Detailed Explanation

This program puts 10 into R0, 20 into R1, then adds them and stores the result in R2.

ARM Architecture
MOV R0, #10
MOV R1, #20
ADD R2, R0, R1
OutputSuccess
Important Notes

Registers R0 to R12 are volatile, meaning their values can change during function calls unless saved.

R13, R14, and R15 have special uses (stack pointer, link register, program counter) and are not general purpose.

Summary

R0 to R12 are general purpose registers used to hold data and addresses.

They help the CPU perform calculations and manage data quickly.

Understanding how to use these registers is key to programming in ARM assembly.

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
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard