Available expressions analysis in Compiler Design - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define an expression as available at a program point if it has been computed and not invalidated.

Compiler Design

An expression is considered available at a point if it has been computed on all paths leading to that point and [1].

Drag options to blanks, or click blank then click option'

Anot modified afterwards

Brecomputed later

Cnever used

Dalways constant

Attempts:

3 left

💡 Hint

Common Mistakes

Assuming an expression is available if it was computed once anywhere.

Ignoring changes to variables used in the expression.

An expression is available if it has been computed and its operands have not been changed since then, so it can be reused safely.

2fill in blank

medium

Complete the code to represent the transfer function for available expressions in a basic block.

Compiler Design

OUT[B] = GEN[B] ∪ (IN[B] [1] KILL[B])

Drag options to blanks, or click blank then click option'

A*

B-

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using union instead of difference for removing killed expressions.

Confusing arithmetic operators with set operations.

The transfer function for available expressions uses set difference to remove killed expressions from the input set before adding generated expressions.

3fill in blank

hard

Fix the error in the data flow equation for IN[B] in available expressions analysis.

Compiler Design

IN[B] = [1](OUT[P]) for all predecessors P of B

Drag options to blanks, or click blank then click option'

Aunion

Bsymmetric_difference

Cdifference

Dintersection

Attempts:

3 left

💡 Hint

Common Mistakes

Using union instead of intersection, which overestimates availability.

Using difference or symmetric difference which are incorrect here.

IN[B] is the intersection of the OUT sets of all predecessors because an expression must be available on all paths to be considered available at B.

4fill in blank

hard

Fill both blanks to complete the iterative algorithm step for available expressions.

Compiler Design

IN[B] = [1](OUT[P]) for all predecessors P of B
OUT[B] = GEN[B] ∪ (IN[B] [2] KILL[B])

Drag options to blanks, or click blank then click option'

Aintersection

Bunion

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing union and intersection in the wrong places.

Using addition instead of set difference.

The iterative step uses intersection for IN[B] to find expressions available on all paths, and set difference to remove killed expressions before adding generated ones for OUT[B].

5fill in blank

hard

Fill all three blanks to define the initial conditions and transfer function for available expressions analysis.

Compiler Design

For entry block E:
IN[E] = [1]
For other blocks B:
IN[B] = [2](OUT[P]) for all predecessors P of B
OUT[B] = GEN[B] ∪ (IN[B] [3] KILL[B])

Drag options to blanks, or click blank then click option'

Aempty set

Bintersection

C-

Duniversal set

Attempts:

3 left

💡 Hint

Common Mistakes

Assuming entry block has all expressions available initially.

Using union instead of intersection for IN[B].

Using addition instead of set difference for OUT[B].

The entry block has no available expressions initially (empty set). For other blocks, IN is the intersection of predecessors' OUT sets. OUT is computed by adding generated expressions to IN minus killed expressions.

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