Common subexpression elimination in Compiler Design - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to identify a common subexpression in the given expressions.

Compiler Design

if (a + b) == [1]:

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Aa * b

Ba - b

Cb + c

Da + b

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing a different expression that is not repeated.

Confusing addition with subtraction or multiplication.

The expression a + b is the common subexpression being checked.

2fill in blank

medium

Complete the code to replace a repeated expression with a temporary variable.

Compiler Design

temp = [1]
result = temp * 2

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Ac * d

Ba - b

Ca + b

Db + c

Attempts:

3 left

💡 Hint

Common Mistakes

Using an expression that is not repeated.

Not assigning the expression to a temporary variable.

The repeated expression a + b is stored in temp to avoid recomputation.

3fill in blank

hard

Fix the error in the code to correctly eliminate the common subexpression.

Compiler Design

result = (x + y) * 2
value = (x + [1]) - 3

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Ay

Bz

Cx

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different variable that breaks the common subexpression.

Replacing with a constant instead of the variable.

The common subexpression is x + y, so y completes the expression.

4fill in blank

hard

Fill both blanks to create code that eliminates the common subexpression 'a * b'.

Compiler Design

[1] = a * b;
result1 = [2] + c;
result2 = [2] * d;

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Atemp

Ba * b

Ca + b

Db * a

Attempts:

3 left

💡 Hint

Common Mistakes

Using different names for the temporary variable.

Putting the expression in the reuse spots instead of the temp name.

The temporary variable temp holds a * b, which is reused in both result1 and result2.

5fill in blank

hard

Fill all three blanks to eliminate common subexpressions in a chained computation.

Compiler Design

[1] = x + y;
[2] = [1] * z;
final = [3] + w;

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At1

Bt2

Dx + y

Attempts:

3 left

💡 Hint

Common Mistakes

Reusing the wrong temporary.

Not chaining the dependencies correctly.

Using the original expressions instead of temps.

t1 = x + y; t2 = t1 * z; final = t2 + w; Reuses t1 and t2 to avoid recomputing subexpressions.