Local optimization (peephole) in Compiler Design - Mini Project: Build & Apply

Local Optimization with Peephole Technique

📖 Scenario: You are working on a simple compiler that translates assembly-like instructions. Sometimes, small sequences of instructions can be improved to run faster or use fewer steps. This is called local optimization using the peephole technique.Imagine you have a short list of instructions, and you want to find and replace inefficient patterns with better ones.

🎯 Goal: Build a small peephole optimizer that scans a list of instructions and replaces a specific inefficient pattern with a better one.For example, replace the sequence LOAD A followed by STORE A with a single MOVE A instruction.

📋 What You'll Learn

Create a list called instructions with the exact sequence: 'LOAD A', 'STORE A', 'ADD B', 'LOAD C', 'STORE C'

Create a variable called optimized_instructions initialized as an empty list

Use a while loop with an index variable i to scan through instructions

Replace every occurrence of 'LOAD A' followed immediately by 'STORE A' with 'MOVE A' in optimized_instructions, copying other instructions unchanged

💡 Why This Matters

🌍 Real World

Compilers use peephole optimization to improve small parts of generated code, making programs run faster and use less memory.

💼 Career

Understanding peephole optimization is important for compiler developers, performance engineers, and anyone working on low-level code optimization.

1

Create the initial instruction list

Create a list called instructions with these exact strings in order: 'LOAD A', 'STORE A', 'ADD B', 'LOAD C', 'STORE C'.

Compiler Design

# Create the list of instructions
# Your code here

Hint

Use square brackets [] to create a list and separate items with commas.

2

Prepare the optimized instructions list and index

Create an empty list called optimized_instructions and a variable i set to 0 to use as an index for scanning.

Compiler Design

instructions = ['LOAD A', 'STORE A', 'ADD B', 'LOAD C', 'STORE C']
# Create an empty list called optimized_instructions
# Create an index variable i set to 0
# Your code here

Hint

Use [] to create an empty list and assign 0 to i.

3

Scan and optimize instructions using a while loop

Use a while loop that runs while i is less than the length of instructions. Inside the loop, check if the current instruction is 'LOAD A' and the next instruction is 'STORE A'. If yes, append 'MOVE A' to optimized_instructions and increase i by 2. Otherwise, append the current instruction and increase i by 1.

Compiler Design

instructions = ['LOAD A', 'STORE A', 'ADD B', 'LOAD C', 'STORE C']
optimized_instructions = []
i = 0
# Use a while loop to scan instructions and optimize
# Your code here

Hint

Remember to check the next instruction exists before accessing it to avoid errors.

4

Complete the peephole optimization

Add a final comment line # Peephole optimization complete after the loop to mark the end of the optimization process.

Compiler Design

instructions = ['LOAD A', 'STORE A', 'ADD B', 'LOAD C', 'STORE C']
optimized_instructions = []
i = 0
while i < len(instructions):
    if i + 1 < len(instructions) and instructions[i] == 'LOAD A' and instructions[i + 1] == 'STORE A':
        optimized_instructions.append('MOVE A')
        i += 2
    else:
        optimized_instructions.append(instructions[i])
        i += 1
# Add a final comment to mark completion
# Your code here

Hint

This comment helps others understand where the optimization ends.

Practice

1. What is the main goal of local optimization (peephole) in compiler design?

easy

A. To improve small sequences of instructions by removing redundancies

B. To rewrite the entire program in a different language

C. To increase the number of instructions for clarity

D. To change the program's output for optimization

Solution

1. Step 1: Understand local optimization scope

   Local optimization focuses on small instruction sequences, not the whole program.

2. Step 2: Identify the goal of peephole optimization

   Peephole optimization removes unnecessary or redundant instructions to make code faster and smaller without changing its meaning.

3. Final Answer:

   To improve small sequences of instructions by removing redundancies -> Option A

4. Quick Check:

   Local optimization = remove redundancies [OK]

Hint: Focus on small instruction sequences and redundancy removal [OK]

Common Mistakes:

• Thinking it rewrites the whole program
• Believing it changes program output
• Assuming it adds more instructions

2. Which of the following is a correct example of a peephole optimization?

easy

A. Replacing LOAD A; LOAD A with LOAD A

B. Changing variable names for readability

C. Adding comments to the code

D. Rearranging functions in the source file

Solution

1. Step 1: Identify peephole optimization examples

   Peephole optimization looks at small instruction sequences to remove redundancies.

2. Step 2: Evaluate each option

   Replacing LOAD A; LOAD A with LOAD A removes duplicate instructions, which is a classic peephole optimization. Options A, B, and D do not optimize instructions.

3. Final Answer:

   Replacing LOAD A; LOAD A with LOAD A -> Option A

4. Quick Check:

   Remove duplicate instructions = peephole optimization [OK]

Hint: Look for removing repeated instructions in short sequences [OK]

Common Mistakes:

• Confusing code readability changes with optimization
• Thinking comments affect optimization
• Assuming rearranging functions is peephole optimization

3. Consider this instruction sequence before peephole optimization:

LOAD R1, #5
ADD R1, R1, #0
STORE R1, 100(R2)

What will be the optimized sequence after applying peephole optimization?

medium

A. LOAD R1, #5
ADD R1, #0
STORE R1, 100(R2)

B. LOAD R1, #5
ADD R1, R1, #0
STORE R1, 100(R2)

C. LOAD R1, #5
STORE R1, 100(R2)

D. ADD R1, R1, #0
STORE R1, 100(R2)

Solution

1. Step 1: Identify redundant instructions

   The instruction ADD R1, R1, #0 adds zero, which does not change the value in R1, so it is redundant.

2. Step 2: Remove redundant instruction

   Removing the redundant ADD instruction leaves LOAD R1, #5 and STORE R1, 100(R2).

3. Final Answer:

   LOAD R1, #5 STORE R1, 100(R2) -> Option C

4. Quick Check:

   Remove add zero = optimized sequence [OK]

Hint: Remove instructions that do not change values (like add zero) [OK]

Common Mistakes:

• Keeping redundant add zero instruction
• Removing load instruction incorrectly
• Misunderstanding instruction effects

4. Identify the error in this peephole optimization attempt:

Original: LOAD R2, #10
SUB R2, R2, #0
STORE R2, 200(R3)

Optimized: LOAD R2, #10
SUB R2, R2, #10
STORE R2, 200(R3)

medium

A. The load instruction should be removed

B. No error, optimization is correct

C. The store instruction is missing

D. The subtraction of zero was incorrectly replaced with subtraction of ten

Solution

1. Step 1: Compare original and optimized subtraction

   Original subtracts zero, which does nothing. Optimized subtracts ten, which changes the value incorrectly.

2. Step 2: Identify the error in optimization

   Replacing subtraction by zero with subtraction by ten changes program meaning, which is incorrect for peephole optimization.

3. Final Answer:

   The subtraction of zero was incorrectly replaced with subtraction of ten -> Option D

4. Quick Check:

   Incorrect instruction change = error in peephole optimization [OK]

Hint: Ensure optimized instructions do not change program meaning [OK]

Common Mistakes:

• Changing instruction values incorrectly
• Removing necessary instructions
• Assuming all changes are optimizations

5. Given this code snippet:

LOAD R3, #0
ADD R3, R3, #5
ADD R3, R3, #0
STORE R3, 300(R4)

Which peephole optimization sequence correctly optimizes this code without changing its meaning?

hard

A. LOAD R3, #0
ADD R3, R3, #0
STORE R3, 300(R4)

B. LOAD R3, #5
STORE R3, 300(R4)

C. ADD R3, R3, #5
STORE R3, 300(R4)

D. LOAD R3, #5
ADD R3, R3, #0
STORE R3, 300(R4)

Solution

1. Step 1: Analyze the original instructions

   Loading zero into R3, then adding 5, then adding 0, is equivalent to loading 5 directly into R3.

2. Step 2: Apply peephole optimization

   Replace the sequence with a single instruction LOAD R3, #5 and then store. This removes redundant instructions without changing meaning.

3. Final Answer:

   LOAD R3, #5 STORE R3, 300(R4) -> Option B

4. Quick Check:

   Combine load and add zero to direct load [OK]

Hint: Combine load and add instructions when possible [OK]

Common Mistakes:

• Keeping redundant add zero
• Not combining load and add
• Removing load instruction incorrectly