Postfix Evaluation

Postfix notation places every operator right after its operands, for example A B C * + for A + B * C. That ordering already encodes exactly which operation happens first, so evaluating it doesn't need any precedence rules at all, just a single left-to-right scan with a stack.

The evaluation algorithm

EvaluatePostfix(postfix)
 
1. Create empty stack S
 
2. For each token in postfix (left to right):
 
   a) If token is an operand:
      Push its value onto S
 
   b) If token is an operator (+, -, *, /):
      operand2 = Pop from S   (this is popped first)
      operand1 = Pop from S   (this is popped second)
      result = operand1 operator operand2
      Push result onto S
 
3. answer = Pop from S (the final result)
4. Return answer

Every operand gets pushed as it's read. Every operator pops the two most recent values, combines them, and pushes the result back, so the stack always holds exactly the intermediate results still waiting to be combined.

Why operand order matters

postfix: A B -
 
Pop B first  → this is the right operand
Pop A second → this is the left operand
Result = A - B, not B - A

Example

Evaluate 5 3 -:

Push 5:  Stack: [5]
Push 3:  Stack: [5, 3]
 
Operator -:
  operand2 = Pop() = 3   (right operand)
  operand1 = Pop() = 5   (left operand)
  result = 5 - 3 = 2
 
Result: 2

Getting the pop order backwards here would silently compute 3 - 5 = -2 instead, the wrong answer, with no error to catch it. This is the single most common mistake in postfix evaluation, worth double-checking every time.

Simple worked example

Evaluate 2 3 * 5 +:

At the end, exactly one value remains on the stack: the final answer, 11.

KTU PYQ: full worked evaluation

This is the second half of the question, evaluating the postfix expression from the Infix to Postfix conversion. Given A=3, B=2, C=4, D=2, E=5, F=8, G=4, substituting into A B C * D / + E F G / - - gives:

3 2 4 * 2 / + 5 8 4 / - -

Final result: 4

Practice

Quick recall

• Postfix evaluation is a single left-to-right scan: push operands, and on each operator pop two values, combine them, and push the result back.
• The first value popped is the right operand, the second value popped is the left operand, this matters for subtraction and division specifically.
• At the end of the scan, the stack holds exactly one value, the result of the whole expression.

That covers every topic in Module 1. For a focused list of the questions most likely to appear from this module, see Most Expected Questions.