Evaluate Postfix Expression Using Stack in DSA C - Execution Trace

Concept Flow - Evaluate Postfix Expression Using Stack

Start

↓

Read next symbol

↓

Is symbol operand?

Yes→Push operand to stack

No↓

Pop two operands from stack

↓

Apply operator

↓

Push result back to stack

↓

More symbols?

Yes→Read next symbol

No↓

Pop final result from stack

↓

End

The flow reads each symbol of the postfix expression, pushes operands to the stack, applies operators by popping two operands, and pushes the result back until the expression ends.

Execution Sample

DSA C

char expr[] = "231*+9-";
int stack[10], top = -1;
for each symbol in expr:
  if operand: push
  else: pop two, apply, push
result = pop()

This code evaluates the postfix expression "231*+9-" step-by-step using a stack.

Execution Table

Step	Operation	Symbol	Stack Before	Action	Stack After	Visual State
1	Read symbol	2	[]	Push 2	[2]	2
2	Read symbol	3	[2]	Push 3	[2,3]	2 -> 3
3	Read symbol	1	[2,3]	Push 1	[2,3,1]	2 -> 3 -> 1
4	Read symbol	*	[2,3,1]	Pop 1 and 3, multiply 3*1=3, push 3	[2,3]	2 -> 3
5	Read symbol	+	[2,3]	Pop 3 and 2, add 2+3=5, push 5	[5]	5
6	Read symbol	9	[5]	Push 9	[5,9]	5 -> 9
7	Read symbol	-	[5,9]	Pop 9 and 5, subtract 5-9=-4, push -4	[-4]	-4
8	End		[-4]	Pop final result	[]	-4

💡 All symbols processed; final result popped from stack.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	Final
top	-1	0	1	2	1	0	1	0	-1
stack	[]	[2]	[2,3]	[2,3,1]	[2,3]	[5]	[5,9]	[-4]	[]

Key Moments - 3 Insights

Why do we pop two operands when we see an operator?

Because postfix operators apply to the two most recent operands on the stack, as shown in steps 4, 5, and 7 in the execution_table.

Why is the first popped operand the right operand and the second popped the left operand?

Because the stack is LIFO, the last pushed operand is the right operand; this order is critical for correct operations like subtraction (step 7).

What happens if the expression is empty or invalid?

The stack will not have enough operands to pop when an operator appears, causing an error; this is why the stack size is checked before popping (implied in the flow).

Visual Quiz - 3 Questions

Look at the execution_table, what is the stack after step 5?

A [2,3,1]

B [2,3]

C [5]

D [3]

💡 Hint

Check the 'Stack After' column in row with Step 5.

At which step does the operator '*' get applied?

A Step 4

B Step 5

C Step 3

D Step 6

💡 Hint

Look for the row where symbol is '*' in execution_table.

If the symbol '9' was missing, what would be the stack after step 6?

A [5,9]

B [5]

C [2,3]

D []

💡 Hint

Refer to the stack state after pushing '9' in step 6 and imagine skipping that step.

Concept Snapshot

Evaluate postfix by reading symbols left to right.
Push operands to stack.
On operator, pop two operands, apply operator, push result.
Repeat until expression ends.
Final stack pop is result.

Full Transcript

This visualization shows how to evaluate a postfix expression using a stack. We read each symbol from left to right. If the symbol is a number, we push it onto the stack. If it is an operator, we pop the top two numbers from the stack, apply the operator, and push the result back. We continue this until all symbols are processed. The final number left on the stack is the result. The example expression "231*+9-" is traced step-by-step, showing stack changes and operations.