Three-address code in Compiler Design - Mini Project: Build & Apply

Understanding Three-Address Code

📖 Scenario: Imagine you are learning how computers translate simple math expressions into a step-by-step list of instructions. This list is called three-address code, which breaks down complex expressions into smaller parts that a computer can easily understand.

🎯 Goal: You will create a simple three-address code representation for the expression a = b + c * d. This will help you see how a computer breaks down and processes this expression step-by-step.

📋 What You'll Learn

Create variables to hold the original expression parts

Add a temporary variable to store intermediate results

Write the three-address code steps using simple assignments

Complete the final assignment to variable a

💡 Why This Matters

🌍 Real World

Three-address code is used inside compilers to translate human-readable code into simpler steps that machines can execute efficiently.

💼 Career

Understanding three-address code is important for compiler developers, software engineers working on low-level programming, and anyone interested in how programming languages work under the hood.

1

DATA SETUP: Define the variables for the expression

Create variables named b, c, and d with values 2, 3, and 4 respectively.

Compiler Design

# Define variables b, c, and d with values 2, 3, and 4
# Your code here

Hint

Use simple assignment statements like b = 2.

2

CONFIGURATION: Create a temporary variable for intermediate result

Create a variable named t1 to hold the result of c * d.

Compiler Design

b = 2
c = 3
d = 4
# Create temporary variable t1 for c * d
# Your code here

Hint

Multiply c and d and store in t1.

3

CORE LOGIC: Compute the sum using the temporary variable

Create a variable named t2 to hold the result of b + t1.

Compiler Design

b = 2
c = 3
d = 4
t1 = c * d
# Create temporary variable t2 for b + t1
# Your code here

Hint

Add b and t1 and store in t2.

4

COMPLETION: Assign the final result to variable a

Assign the value of t2 to variable a to complete the expression.

Compiler Design

b = 2
c = 3
d = 4
t1 = c * d
t2 = b + t1
# Assign t2 to a
# Your code here

Hint

Complete the expression by assigning t2 to a.

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