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DSA Cprogramming~10 mins

Evaluate Postfix Expression Using Stack in DSA C - Execution Trace

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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
StepOperationSymbolStack BeforeActionStack AfterVisual State
1Read symbol2[]Push 2[2]2
2Read symbol3[2]Push 3[2,3]2 -> 3
3Read symbol1[2,3]Push 1[2,3,1]2 -> 3 -> 1
4Read symbol*[2,3,1]Pop 1 and 3, multiply 3*1=3, push 3[2,3]2 -> 3
5Read symbol+[2,3]Pop 3 and 2, add 2+3=5, push 5[5]5
6Read symbol9[5]Push 9[5,9]5 -> 9
7Read symbol-[5,9]Pop 9 and 5, subtract 5-9=-4, push -4[-4]-4
8End[-4]Pop final result[]-4
💡 All symbols processed; final result popped from stack.
Variable Tracker
VariableStartAfter 1After 2After 3After 4After 5After 6After 7Final
top-10121010-1
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
Test your understanding
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?
AStep 4
BStep 5
CStep 3
DStep 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.

Practice

(1/5)
1. What is the main data structure used to evaluate a postfix expression?
easy
A. Stack
B. Queue
C. Array
D. Linked List

Solution

  1. Step 1: Understand postfix evaluation process

    Postfix expressions are evaluated by scanning from left to right and using a data structure to store operands temporarily.
  2. Step 2: Identify the suitable data structure

    A stack is used because it allows pushing operands and popping them when an operator is encountered, following Last-In-First-Out order.
  3. Final Answer:

    Stack -> Option A
  4. Quick Check:

    Postfix evaluation uses stack = Stack [OK]
Hint: Remember: postfix needs LIFO, so use stack [OK]
Common Mistakes:
  • Confusing stack with queue
  • Using array without stack operations
  • Thinking linked list is primary for postfix
2. Which of the following is the correct syntax to push an integer value 'x' onto a stack named 's' in C?
easy
A. stack_push(x);
B. s.push(x);
C. push(x, s);
D. push(s, x);

Solution

  1. Step 1: Understand typical stack push function signature

    In C, stack push usually takes the stack and the value to push as arguments, e.g., push(stack, value).
  2. Step 2: Match the correct syntax

    push(s, x); matches the common pattern push(s, x); where 's' is the stack and 'x' is the value.
  3. Final Answer:

    push(s, x); -> Option D
  4. Quick Check:

    push(stack, value) = push(s, x) [OK]
Hint: Remember: push(stack, value) in C [OK]
Common Mistakes:
  • Using object.method() syntax in C
  • Swapping argument order
  • Using undefined function names
3. What is the result of evaluating the postfix expression "6 2 3 + -" using a stack?
medium
A. 3
B. 1
C. 5
D. -1

Solution

  1. Step 1: Evaluate postfix step-by-step

    Push 6, push 2, push 3. Encounter '+', pop 3 and 2, add: 2+3=5, push 5. Encounter '-', pop 5 and 6, subtract: 6-5=1.
  2. Step 2: Verify final stack value

    After all operations, the stack has one value: 1.
  3. Final Answer:

    1 -> Option B
  4. Quick Check:

    6 - (2 + 3) = 1 [OK]
Hint: Process operators by popping two operands [OK]
Common Mistakes:
  • Subtracting in wrong order (operand2 - operand1)
  • Adding instead of subtracting
  • Not pushing intermediate results
4. Identify the error in this C code snippet for postfix evaluation:
char* expr = "5 9 8 + 4 6 * * 7 + *";
int i = 0;
while(expr[i] != '\0') {
  if(isdigit(expr[i])) push(stack, expr[i]);
  else { /* operator handling */ }
  i++;
}
medium
A. Pushing character digit instead of integer value
B. Using while loop instead of for loop
C. Missing stack initialization
D. Incorrect operator handling

Solution

  1. Step 1: Check what is pushed onto stack

    The code pushes expr[i], which is a character like '5', not the integer 5.
  2. Step 2: Explain why this is wrong

    Characters have ASCII codes; pushing '5' pushes 53, not integer 5, causing wrong calculations.
  3. Final Answer:

    Pushing character digit instead of integer value -> Option A
  4. Quick Check:

    Push integer, not char digit [OK]
Hint: Convert char to int before push: expr[i] - '0' [OK]
Common Mistakes:
  • Pushing char instead of int
  • Ignoring spaces in expression
  • Not handling operators properly
5. Given the postfix expression "4 5 7 2 + - *", what is the final evaluated result using stack evaluation?
hard
A. -24
B. 16
C. -16
D. 24

Solution

  1. Step 1: Evaluate inner expression 7 2 +

    Push 4, 5, 7, 2. '+' operator: pop 2 and 7, add 7+2=9, push 9.
  2. Step 2: Evaluate next operator '-'

    Next operator '-', pop 9 and 5, subtract 5-9 = -4, push -4.
  3. Step 3: Evaluate final operator '*'

    Pop -4 and 4, multiply 4 * -4 = -16, push -16.
  4. Final Answer:

    -16 -> Option C
  5. Quick Check:

    4 * (5 - (7 + 2)) = -16 [OK]
Hint: Evaluate inner expressions first, then combine [OK]
Common Mistakes:
  • Wrong operator order in subtraction
  • Not evaluating nested operations correctly
  • Mixing up operand order in multiplication
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