Stack Basics
Basic Stack Operations in C++
C++ Stack Basics Program
This program helps you to learn the fundamental structure and syntax of C++ programming.
#include <iostream>
#include <stack>
using namespace std;
int main() {
// Create stack
stack<int> st;
// Push elements
st.push(10);
st.push(20);
st.push(30);
st.push(40);
st.push(50);
cout << "Stack size: " << st.size() << endl;
cout << "Top element: " << st.top() << endl;
// Display stack (by popping)
cout << "\nStack elements (LIFO - Last In First Out):" << endl;
while (!st.empty()) {
cout << st.top() << " ";
st.pop();
}
cout << endl;
// Stack for string reversal
stack<char> charStack;
string str = "Hello";
cout << "\nOriginal string: " << str << endl;
// Push all characters
for (char c : str) {
charStack.push(c);
}
// Pop to reverse
string reversed = "";
while (!charStack.empty()) {
reversed += charStack.top();
charStack.pop();
}
cout << "Reversed string: " << reversed << endl;
// Check balanced parentheses
stack<char> parenStack;
string expression = "((()))";
bool balanced = true;
for (char c : expression) {
if (c == '(') {
parenStack.push(c);
} else if (c == ')') {
if (parenStack.empty()) {
balanced = false;
break;
}
parenStack.pop();
}
}
if (parenStack.empty() && balanced) {
cout << "\nExpression '" << expression << "' is balanced" << endl;
} else {
cout << "\nExpression '" << expression << "' is not balanced" << endl;
}
return 0;
}Stack size: 5 Top element: 50 Stack elements (LIFO - Last In First Out): 50 40 30 20 10 Original string: Hello Reversed string: olleH Expression '((()))' is balanced
Understanding Stack Basics
This program teaches you how to use Stack Basics in C++. Stack is a LIFO (Last In First Out) container where elements are added and removed from the top. It's perfect for problems requiring reverse order processing, expression evaluation, and maintaining state.
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1. What This Program Does
The program demonstrates stack operations:
Stacks provide simple, efficient last-in-first-out data access.
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2. Header Files Used
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3. Understanding Stack
LIFO Concept
:
Key Features
:
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4. Stack Operations
push()
:
st.push(10); // Add to top
pop()
:
st.pop(); // Remove from top
top()
:
int value = st.top(); // Access top element
How it works
:
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5. Stack Properties
Size and Empty
:
st.size() // Number of elements
st.empty() // True if empty
Access
:
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6. String Reversal
Using Stack
:
How it works
:
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7. Balanced Parentheses
Algorithm
:
How it works
:
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8. When to Use Stack
Best For
:
Example Scenarios
:
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9. Important Considerations
LIFO Order
:
Performance
:
Limitations
:
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10. return 0;
This ends the program successfully.
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Summary
This program is fundamental for learning stack data structure, understanding LIFO operations, and preparing for expression evaluation and algorithm implementation in C++ programs.
Let us now understand every line and the components of the above program.
Note: To write and run C++ programs, you need to set up the local environment on your computer. Refer to the complete article Setting up C++ Development Environment. If you do not want to set up the local environment on your computer, you can also use online IDE to write and run your C++ programs.
Practical Learning Notes for Stack Basics
This C++ program is part of the "STL Containers Programs" topic and is designed to help you build real problem-solving confidence, not just memorize syntax. Start by understanding the goal of the program in plain language, then trace the logic line by line with a custom input of your own. Once you can predict the output before running the code, your understanding becomes much stronger.
A reliable practice pattern is to run the original version first, then modify only one condition or variable at a time. Observe how that single change affects control flow and output. This deliberate style helps you understand loops, conditions, and data movement much faster than copying full solutions repeatedly.
For interview preparation, explain this solution in three layers: the high-level approach, the step-by-step execution, and the time-space tradeoff. If you can teach these three layers clearly, you are ready to solve close variations of this problem under time pressure.