Three-address code in Compiler Design - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to represent the assignment in three-address code.

Compiler Design

t1 = [1]

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

Aa / b

Ba * b

Ca - b

Da + b

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication or division instead of addition.

Forgetting to assign the result to a temporary variable.

The three-address code for assigning the sum of a and b to a temporary variable t1 is t1 = a + b.

2fill in blank

medium

Complete the code to represent the conditional jump in three-address code.

Compiler Design

if [1] goto L1

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

Aa > b

Ba = b

Ca < b

Da != b

Attempts:

3 left

💡 Hint

Common Mistakes

Using equality or inequality instead of greater than.

Using assignment operator instead of comparison.

The conditional jump checks if a is greater than b. If true, it jumps to label L1.

3fill in blank

hard

Fix the error in the three-address code for multiplication.

Compiler Design

t2 = [1]

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

Aa + b

Ba / b

Ca * b

Da - b

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition or subtraction instead of multiplication.

Using division operator mistakenly.

The correct three-address code for multiplication assigns a * b to t2.

4fill in blank

hard

Fill both blanks to complete the three-address code for a conditional jump.

Compiler Design

if [1] [2] y goto L2

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

Ax

By

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the variables or using the wrong operator.

Using assignment operator instead of relational operator.

The condition checks if x is greater than y. If true, it jumps to label L2.

5fill in blank

hard

Fill all three blanks to complete the three-address code for an assignment.

Compiler Design

t3 = [1] [2] [3]

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

Aa

B+

Cb

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition.

Swapping the order of variables.

The three-address code assigns the sum of a and b to t3.

Practice

1. What is the main purpose of three-address code in compiler design?

easy

A. To execute machine code directly on hardware

B. To break complex instructions into simpler steps with at most three parts

C. To store source code in a compressed format

D. To replace high-level language syntax with natural language

Solution

1. Step 1: Understand the role of intermediate code

   Intermediate code like three-address code simplifies complex instructions for easier processing.

2. Step 2: Identify the key feature of three-address code

   It breaks instructions into steps with at most three parts (two operands and one result).

3. Final Answer:

   To break complex instructions into simpler steps with at most three parts -> Option B

4. Quick Check:

   Three-address code = simple 3-part instructions [OK]

Hint: Three-address code simplifies instructions into three parts [OK]

Common Mistakes:

• Confusing three-address code with machine code
• Thinking it compresses source code
• Believing it replaces programming languages

2. Which of the following is a correct example of a three-address code instruction?

easy

A. if a > b goto L1 else goto L2

B. while (a < b) { a = a + 1; }

C. print(a, b, c)

D. t1 = a + b

Solution

1. Step 1: Identify the format of three-address code

   It typically has one operator and up to two operands, assigning result to a temporary variable.

2. Step 2: Check each option

   t1 = a + b matches the format: one operator (+), two operands (a, b), and a result (t1). Others are control flow or high-level statements.

3. Final Answer:

   t1 = a + b -> Option D

4. Quick Check:

   Three-address code = one operator, two operands, one result [OK]

Hint: Look for one operator and two operands with a result variable [OK]

Common Mistakes:

• Choosing control flow statements as three-address code
• Confusing function calls with three-address code
• Ignoring the assignment to a temporary variable

3. Given the three-address code:
t1 = a + b
t2 = t1 * c
d = t2 - e
What is the final value assigned to d in terms of a, b, c, e?

medium

A. (a + b) * c - e

B. a + b * c - e

C. a + (b * c) - e

D. (a + b) - (c * e)

Solution

1. Step 1: Evaluate intermediate expressions step-by-step

   t1 = a + b; then t2 = t1 * c = (a + b) * c; finally d = t2 - e = (a + b) * c - e.

2. Step 2: Substitute and simplify

   The final expression for d is (a + b) * c - e, matching (a + b) * c - e.

3. Final Answer:

   (a + b) * c - e -> Option A

4. Quick Check:

   Follow code stepwise = (a + b) * c - e [OK]

Hint: Follow temporary variables stepwise to build final expression [OK]

Common Mistakes:

• Ignoring parentheses changing operation order
• Multiplying b by c before adding a
• Mixing subtraction order

4. Identify the error in this three-address code snippet:
t1 = a + b
t2 = t1 *
d = t2 - e

medium

A. Missing operand after '*' operator in second line

B. Incorrect use of temporary variables

C. Assignment to 'd' should be before 't2'

D. No error, code is correct

Solution

1. Step 1: Examine the second line syntax

   The second line ends with '*' operator but lacks the second operand after '*', which is invalid.

2. Step 2: Confirm other lines are correct

   First and third lines have proper assignments and operands; only second line has syntax error.

3. Final Answer:

   Missing operand after '*' operator in second line -> Option A

4. Quick Check:

   Operator must have two operands [OK]

Hint: Check each operator has two operands in three-address code [OK]

Common Mistakes:

• Ignoring incomplete operator usage
• Thinking variable order causes error
• Assuming code is correct without checking syntax

5. How would you represent the high-level statement if (a < b) c = a + b; else c = a - b; in three-address code?

hard

A. if a < b then c = a + b else c = a - b

B. c = a + b if a < b else c = a - b

C. if a < b goto L1c = a - bgoto L2L1: c = a + bL2:

D. t1 = a < bc = t1 ? a + b : a - b

Solution

1. Step 1: Translate conditional jump

   Use 'if a < b goto L1' to jump to true branch, else execute false branch.

2. Step 2: Write false branch, jump over true branch

   False branch: c = a - b; then jump to L2 to skip true branch.

3. Step 3: Write true branch and label

   Label L1: c = a + b; Label L2 marks end of conditional.

4. Final Answer:

   if a < b goto L1
   c = a - b
   goto L2
   L1: c = a + b
   L2: -> Option C

5. Quick Check:

   Use labels and jumps for if-else in three-address code [OK]

Hint: Use labels and jumps to represent if-else in three-address code [OK]

Common Mistakes:

• Writing high-level if-else syntax directly
• Using ternary operator not valid in three-address code
• Missing jump to skip true branch after false branch