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].

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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])

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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

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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])

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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])

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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.