### Understanding the Output of a C++ Program#### Problem Statement**Question:** What is the output of the following C++ code if the input is `"1 2 3"`?```cpp#include using namespace std;int main(){ int num1, num2, num3; double average; cin >> num1 >> num2 >> num3; average = (num1 + num2 + num3) / 3.0; cout << average << endl; cin.ignore(); return 0;}```#### Explanation1. **Include Directive**: - `#include ` is used to include the standard input-output stream library, which allows the use of `cin` and `cout`.2. **Namespace**: - `using namespace std;` allows you to use all elements in the `std` namespace without prefixing them with `std::`.3. **Variable Declaration**: - `int num1, num2, num3;` declares three integer variables. - `double average;` declares a double variable to store the average.4. **Input Operations**: - `cin >> num1 >> num2 >> num3;` takes three integer inputs from the user and assigns them to `num1`, `num2`, and `num3`.5. **Calculation**: - `average = (num1 + num2 + num3) / 3.0;` calculates the average of the three numbers. The division is done using a floating-point number (`3.0`) to ensure the result is a double.6. **Output**: - `cout << average << endl;` outputs the computed average followed by a newline.7. **Program Termination**: - `cin.ignore();` is used to ignore any additional input or newline characters. - `return 0;` signifies the successful completion of the program.#### Sample OutputFor the input `"1 2 3"`, the program will compute the average as:\[ \text{average} = \frac{1 + 2 + 3}{3.0} = 2.0 \]**Output:** `2` (formatted as a floating-point number, but displayed without decimal places since the `.0` is not shown by default with standard output settings)

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Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

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What is the output of the following code if the input is "1 2 3": #include using namespace std; int main(){ int num1, num2, num3; double average; cin >> num1 >» num2 >> num3; average = (num1 + num2 + num3) / 3.0; cout <« average <« endl; cin.ignore(); return e; }