Three-address code in Compiler Design - Step-by-Step Execution

Concept Flow - Three-address code

Start with high-level expression

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Break expression into simple operations

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Assign each operation to a temporary variable

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Generate three-address instructions

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Use temporaries for next operations

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Repeat until full expression is converted

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End

The process breaks a complex expression into simple steps, each with at most three parts: two operands and one result, assigned to temporary variables.

Execution Sample

Compiler Design

a = b + c * d

Convert the expression into three-address code by breaking it into simpler steps with temporary variables.

Analysis Table

Step	Operation	Operands	Temporary Variable	Resulting Instruction
1	Multiply	c, d	t1	t1 = c * d
2	Add	b, t1	t2	t2 = b + t1
3	Assign	t2	a	a = t2
4	End	-	-	-

💡 All parts of the expression are converted into simple three-address instructions.

State Tracker

Variable	Start	After Step 1	After Step 2	After Step 3
t1	-	c * d	c * d	c * d
t2	-	-	b + t1	b + t1
a	-	-	-	b + (c * d)

Key Insights - 3 Insights

Why do we need temporary variables like t1 and t2?

Temporary variables hold intermediate results so each instruction has at most two operands and one result, as shown in steps 1 and 2 of the execution_table.

Why can't we write the whole expression in one instruction?

Three-address code limits each instruction to one operation for simplicity and easier optimization, so complex expressions are split into multiple steps as seen in the execution_table.

What does the 'Assign' operation do in step 3?

It stores the final computed value from temporary variable t2 into the original variable a, completing the translation.

Visual Quiz - 3 Questions

Look at the execution_table, what is the operation performed at step 2?

A Assign t2 to a

B Multiply c and d

C Add b and t1

D End of instructions

💡 Hint

Check the 'Operation' and 'Operands' columns in row 2 of the execution_table.

At which step is the temporary variable t1 created and what does it store?

A Step 1, stores c * d

B Step 1, stores b + c

C Step 2, stores b + t1

D Step 3, stores final result

💡 Hint

Look at the 'Temporary Variable' and 'Operation' columns in step 1 of the execution_table.

If the expression was a = (b + c) * d, how would the first temporary variable change?

A t1 = b * d

B t1 = b + c

C t1 = c * d

D t1 = b + d

💡 Hint

Remember the order of operations and how the expression is broken down into steps.

Concept Snapshot

Three-address code breaks complex expressions into simple steps.
Each instruction has at most two operands and one result.
Temporary variables hold intermediate results.
This makes code easier to optimize and translate.
Example: a = b + c * d becomes t1 = c * d; t2 = b + t1; a = t2.

Full Transcript

Three-address code is a way to simplify complex expressions in programming languages during compilation. It breaks down expressions into simple instructions with at most two operands and one result. For example, the expression a = b + c * d is split into steps: first multiply c and d storing in t1, then add b and t1 storing in t2, and finally assign t2 to a. Temporary variables like t1 and t2 hold intermediate results. This step-by-step breakdown helps compilers optimize and generate machine code more easily.

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