Available expressions analysis in Compiler Design - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is Available Expressions Analysis in compiler design?

It is a data flow analysis technique used to find expressions that have already been computed and whose values are still available at a given point in the program, so they can be reused without recomputation.

beginner

Why is Available Expressions Analysis useful?

It helps optimize programs by identifying expressions that do not need to be recalculated, enabling the compiler to reuse previous results and improve efficiency.

intermediate

What does it mean if an expression is available at a program point?

It means the expression has been computed earlier along every path to that point, and none of its variables have changed since then.

intermediate

How does Available Expressions Analysis handle variable assignments?

When a variable involved in an expression is assigned a new value, that expression is no longer available because its previous value might be outdated.

beginner

What is the direction of data flow in Available Expressions Analysis?

It is a forward data flow analysis because it propagates information from the start of the program towards the end.

What does Available Expressions Analysis identify in a program?

A Expressions already computed and unchanged

B Variables that are never used

C Functions that are never called

D Loops that never terminate

Available Expressions Analysis finds expressions that have been computed and whose values are still valid.

When is an expression NOT considered available at a program point?

A If any variable in the expression was reassigned

B If it was never computed before

C If the expression is inside a loop

D If the program has a function call

Reassignment of any variable in the expression invalidates its availability.

What type of data flow analysis is Available Expressions Analysis?

A Backward

B Bidirectional

C Forward

D None of the above

It is a forward analysis because it moves from the start to the end of the program.

Which optimization can benefit directly from Available Expressions Analysis?

A Loop unrolling

B Common subexpression elimination

C Dead code elimination

D Inlining functions

Common subexpression elimination uses available expressions to avoid recomputing the same expression.

If an expression is available at a program point, what can the compiler do?

A Recompute the expression

B Delete the expression

C Ignore the expression

D Reuse the previously computed value

The compiler can reuse the value to save computation time.

Explain what Available Expressions Analysis is and why it is important in compiler optimization.

Think about expressions that can be reused without recalculating.

Describe how variable assignments affect the availability of expressions in Available Expressions Analysis.

Consider what happens when a variable changes after an expression was computed.

Practice

1. What is the main purpose of available expressions analysis in compiler design?

easy

A. To find expressions that have already been computed and not changed

B. To detect syntax errors in the source code

C. To allocate memory for variables

D. To generate machine code from source code

Solution

1. Step 1: Understand the goal of available expressions analysis

   This analysis identifies expressions that have been computed before and whose values have not changed since.

2. Step 2: Compare with other compiler tasks

   Unlike syntax checking or code generation, available expressions focus on optimization by reusing computed values.

3. Final Answer:

   To find expressions that have already been computed and not changed -> Option A

4. Quick Check:

   Available expressions = computed and unchanged expressions [OK]

Hint: Focus on expressions reused without recomputation [OK]

Common Mistakes:

• Confusing with syntax error detection
• Thinking it allocates memory
• Mixing with code generation tasks

2. Which of the following correctly represents the GEN set in available expressions analysis?

easy

A. All variables used in the program

B. Variables assigned new values in a block

C. Expressions computed in a block and not killed within it

D. Expressions that are never computed in the program

Solution

1. Step 1: Define the GEN set

   GEN set contains expressions that are computed in a block and remain valid (not overwritten) within that block.

2. Step 2: Eliminate incorrect options

   Variables assigned new values relate to KILL sets, not GEN. Expressions never computed or all variables are unrelated.

3. Final Answer:

   Expressions computed in a block and not killed within it -> Option C

4. Quick Check:

   GEN = computed and valid expressions [OK]

Hint: GEN = expressions generated and still valid in block [OK]

Common Mistakes:

• Mixing GEN with KILL sets
• Confusing variables with expressions
• Selecting unrelated program elements

3. Consider a basic block with the following statements:

a = b + c
d = a + e
b = 5

Which expressions are available at the end of this block?

medium

A. {a + e}

B. {} (empty set)

C. {b + c, a + e}

D. {b + c}

Solution

1. Step 1: Identify expressions computed

   The expressions computed are b + c and a + e.

2. Step 2: Check if expressions remain valid

   b + c is killed because b is assigned later (b = 5). a + e remains available since neither a nor e is reassigned after its computation.

3. Final Answer:

   {a + e} -> Option A

4. Quick Check:

   Expressions available if no operand reassigned after computation [OK]

Hint: Check operand reassignments after each computation [OK]

Common Mistakes:

• Ignoring variable reassignment kills expressions
• Assuming all computed expressions remain available
• Confusing direct operands with transitive dependencies

4. Identify the error in this available expressions analysis step:

GEN = {x + y}
KILL = {x = 3}
Available expressions at block end = GEN ∪ (IN - KILL)

What is wrong with this calculation?

medium

A. Using union (∪) instead of intersection (∩) for available expressions

B. Confusing KILL set with GEN set

C. Not updating IN set before calculation

D. Including variables instead of expressions in KILL

Solution

1. Step 1: Recall available expressions formula

   The formula Available_out = GEN ∪ (IN - KILL) is correct when GEN and KILL contain expressions.

2. Step 2: Identify the error in the sets

   KILL = {x = 3} includes an assignment instead of killed expressions (e.g., those with x as operand).

3. Final Answer:

   Including variables instead of expressions in KILL -> Option D

4. Quick Check:

   KILL = expressions killed by assignments [OK]

Hint: KILL contains killed expressions, not assignments [OK]

Common Mistakes:

• Misrepresenting KILL set contents
• Mixing variables and expressions in sets
• Ignoring proper IN set update

5. Given two basic blocks with:

Block 1: GEN = {a + b}, KILL = {a = 4}
Block 2: GEN = {a + b, c + d}, KILL = {b = 2}

If the IN set for Block 2 is {a + b}, what is the OUT set for Block 2 after available expressions analysis?

hard

A. {a + b, c + d}

B. {c + d}

C. {a + b}

D. {} (empty set)

Solution

1. Step 1: Calculate OUT set formula

   OUT = GEN ∪ (IN ∩ ¬KILL). Here, GEN = {a + b, c + d}, IN = {a + b}, KILL kills expressions involving b.

2. Step 2: Remove killed expressions from IN

   a + b is killed (involves b), so IN ∩ ¬KILL = {}.

3. Step 3: Compute OUT set

   OUT = GEN ∪ {} = {a + b, c + d}, but since a + b is killed by KILL, it should not be in OUT. Therefore, OUT = {c + d}.

4. Final Answer:

   {c + d} -> Option B

5. Quick Check:

   OUT = GEN ∪ (filtered IN) with killed expressions removed [OK]

Hint: Filter IN with ¬KILL, then union GEN [OK]

Common Mistakes:

• Applying KILL to GEN expressions
• Not intersecting IN with complement of KILL
• Assuming all GEN expressions remain available