Local optimization (peephole) in Compiler Design - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

What is the primary goal of peephole optimization in compilers?

Peephole optimization is a technique used in compilers. What is its main purpose?

A To optimize small sequences of instructions locally for better performance or size

B To improve the overall program logic by changing algorithms

C To translate high-level code directly into machine code

D To perform syntax checking and error reporting

💡 Hint

Think about small code segments and local improvements.

Peephole optimization focuses on small windows of instructions to replace inefficient patterns with better ones, improving speed or reducing code size without changing program behavior.

📋 Factual

intermediate

Which of the following is a common peephole optimization technique?

Identify the typical peephole optimization from the options below.

A Loop unrolling to reduce loop overhead

B Inlining functions to avoid call overhead

C Replacing a sequence of instructions with a single equivalent instruction

D Register allocation across the entire program

💡 Hint

Focus on small instruction sequences and their replacement.

Peephole optimization often replaces multiple instructions with a single, more efficient instruction to improve performance or reduce code size.

🔍 Analysis

advanced

What is the output of the peephole optimizer on this instruction sequence?

Given the following assembly instructions, what will a peephole optimizer most likely produce?

Compiler Design

MOV R1, R2
MOV R2, R1
ADD R1, #0

A

MOV R1, R2
MOV R2, R1
ADD R1, #0

B MOV R1, R2

C

NOP
NOP
NOP

D

MOV R1, R2
NOP

💡 Hint

Consider redundant moves and adding zero.

The second MOV reverses the first, so both cancel out. Adding zero does nothing. The optimizer removes redundant instructions, leaving only the effective MOV.

❓ Reasoning

advanced

Why might peephole optimization sometimes be limited in improving program performance?

Choose the best explanation for the limitations of peephole optimization.

A It increases the size of the compiled code

B It only works on high-level source code, not machine code

C It requires manual intervention by the programmer

D It focuses on small instruction windows and cannot optimize across larger code structures

💡 Hint

Think about the scope of peephole optimization.

Peephole optimization looks at small groups of instructions and cannot analyze or optimize larger program structures or algorithms, limiting its impact.

❓ Comparison

expert

Compare peephole optimization with global optimization in compilers.

Which statement correctly contrasts peephole optimization with global optimization?

A Peephole optimization analyzes entire programs, while global optimization only looks at small instruction sequences

B Global optimization can optimize across functions and modules, whereas peephole optimization only works on local instruction windows

C Peephole optimization requires source code changes, global optimization does not

D Global optimization is faster and simpler than peephole optimization

💡 Hint

Consider the scope and scale of each optimization type.

Peephole optimization works on small, local instruction sequences, while global optimization analyzes larger program parts, including multiple functions or modules, to improve performance more broadly.

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