General purpose registers (R0-R12) in ARM Architecture - Step-by-Step Execution

Concept Flow - General purpose registers (R0-R12)

Start Instruction

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Fetch Instruction

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Decode Instruction

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Identify Registers R0-R12

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Read/Write Data in Registers

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Execute Operation

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Store Result in Register

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Next Instruction or End

This flow shows how ARM instructions use general purpose registers R0 to R12 to hold data during execution.

Execution Sample

ARM Architecture

MOV R0, #5
MOV R1, #10
ADD R2, R0, R1
SUB R3, R2, #3

This code moves values into registers R0 and R1, adds them into R2, then subtracts 3 from R2 storing result in R3.

Analysis Table

Step	Instruction	Registers Before	Operation	Registers After
1	MOV R0, #5	R0=0, R1=0, R2=0, R3=0	Set R0 to 5	R0=5, R1=0, R2=0, R3=0
2	MOV R1, #10	R0=5, R1=0, R2=0, R3=0	Set R1 to 10	R0=5, R1=10, R2=0, R3=0
3	ADD R2, R0, R1	R0=5, R1=10, R2=0, R3=0	Add R0 + R1, store in R2	R0=5, R1=10, R2=15, R3=0
4	SUB R3, R2, #3	R0=5, R1=10, R2=15, R3=0	Subtract 3 from R2, store in R3	R0=5, R1=10, R2=15, R3=12

💡 All instructions executed, final register values updated.

State Tracker

Register	Start	After Step 1	After Step 2	After Step 3	After Step 4
R0	0	5	5	5	5
R1	0	0	10	10	10
R2	0	0	0	15	15
R3	0	0	0	0	12

Key Insights - 3 Insights

Why does R2 remain 0 after step 2 even though R0 and R1 have values?

Because step 2 only sets R1; R2 is updated only in step 3 when the ADD instruction executes (see execution_table row 3).

Why can we use immediate values like #5 and #10 in MOV instructions?

MOV allows loading a constant number directly into a register, so #5 and #10 are immediate values assigned to R0 and R1 respectively (see execution_table rows 1 and 2).

What happens if we try to use a register outside R0-R12 here?

This example focuses on R0-R12 which are general purpose; other registers like R13 (stack pointer) have special roles and are not used for general data storage in this context.

Visual Quiz - 3 Questions

Look at the execution table at step 3. What is the value of R2 after the ADD instruction?

A 5

B 10

C 15

D 0

💡 Hint

Check the 'Registers After' column in row 3 of the execution_table.

At which step does R3 get its first non-zero value?

A Step 2

B Step 4

C Step 3

D Step 1

💡 Hint

Look at the 'Registers After' column for R3 in the execution_table rows.

If the instruction 'SUB R3, R2, #3' was changed to 'SUB R3, R1, #3', what would be the value of R3 after step 4?

A 7

B 12

C 15

D 10

💡 Hint

Subtract 3 from R1's value after step 2 (see variable_tracker).

Concept Snapshot

General purpose registers R0-R12 hold data for ARM instructions.
MOV loads values into registers.
ADD and SUB perform arithmetic using registers.
Registers store intermediate and final results.
R0-R12 are used for general data operations.

Full Transcript

This visual execution trace shows how ARM general purpose registers R0 to R12 are used in simple instructions. We start with all registers at zero. The MOV instructions load immediate values 5 and 10 into R0 and R1. Then ADD adds R0 and R1, storing 15 in R2. Finally, SUB subtracts 3 from R2, storing 12 in R3. The execution table tracks register values before and after each step. The variable tracker shows how each register changes over time. Key moments clarify common confusions about when registers update and the use of immediate values. The quiz tests understanding of register values at specific steps and effects of instruction changes. This helps beginners see how data moves and changes inside ARM CPU registers during execution.

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