# C++ Program to Find Largest Number Among Three Numbers

## Source Code

``````#include <iostream>

using namespace std;

int main()
{
int num1, num2, num3; // Declare variables to hold the three numbers

cout << "Enter the first number: "; // Prompt the user to enter the first number
cin >> num1; // Read the first number from the standard input stream

cout << "Enter the second number: "; // Prompt the user to enter the second number
cin >> num2; // Read the second number from the standard input stream

cout << "Enter the third number: "; // Prompt the user to enter the third number
cin >> num3; // Read the third number from the standard input stream

int largest = num1; // Initialize the largest number to the first number

// Compare the second number to the largest number
if (num2 > largest)
largest = num2;

// Compare the third number to the largest number
if (num3 > largest)
largest = num3;

cout << "The largest number is: " << largest << endl; // Print the largest number

return 0;
}
``````

## Code Explanation

This program declares three variables `num1`, `num2`, and `num3` of type `int` to hold the three numbers entered by the user. It then uses the `cout` object to print prompts asking the user to enter the numbers and the `cin` object to read the numbers from the standard input stream.

To find the largest number, the program initializes a variable `largest` to the value of `num1`. It then uses two `if` statements to compare `num2` and `num3` to `largest` and update `largest` if either of these numbers is greater than `largest`.

Finally, the program uses the `cout` object to print the value of `largest`, which is the largest number among the three numbers.

In C++, the `if-else` statement is a control flow statement that allows you to execute a block of code conditionally.
The `if` clause specifies a condition, and the code inside the `{}` brackets following the `if` clause is executed if the condition is true. The `else` clause specifies an alternative block of code to execute if the condition is false.