What will the output be after the execution of this code? void changeToFive(int num1); int main() { int num = 1; cout << "In the main function, the number is: " << num << endl; changeToFive(num); cout << "Now the output is " << num << endl; return 0; } void changeToFive(int num1) { num1 = 5; cout << "The number is now: << num1 << endl; }

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
Section: Chapter Questions
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### Code Explanation

In this example, we have a C++ code snippet that involves a function call and outputs to the console. Let's break it down:

#### Code
```cpp
void changeToFive(int num1);
int main()
{
    int num = 1;
    cout << "In the main function, the number is: " << num << endl;
    changeToFive(num);
    cout << "Now the output is " << num << endl;
    return 0;
}

void changeToFive(int num1)
{
    num1 = 5;
    cout << "The number is now: " << num1 << endl;
}
```

#### Analysis

1. **Main Function Execution:**
   - The program begins by entering the `main()` function.
   - An integer variable `num` is initialized with the value `1`.
   - It prints: `"In the main function, the number is: 1"`

2. **Function Call `changeToFive(num)`:**
   - The value of `num` (which is `1`) is passed to the `changeToFive` function.
   - Inside `changeToFive`, the variable `num1` is a copy of `num`.
   - The line `num1 = 5;` changes the local copy `num1` to `5`.
   - It prints: `"The number is now: 5"`
   - Note: This change does not affect `num` in the `main()` function.

3. **After the Function Call:**
   - Control returns to `main()`.
   - It prints: `"Now the output is 1"`

#### Conclusion

The output after executing this code will be:

```
In the main function, the number is: 1
The number is now: 5
Now the output is 1
```

This illustrates the concept of pass-by-value in C++, where modifications to function parameters do not affect the original arguments.
Transcribed Image Text:### Code Explanation In this example, we have a C++ code snippet that involves a function call and outputs to the console. Let's break it down: #### Code ```cpp void changeToFive(int num1); int main() { int num = 1; cout << "In the main function, the number is: " << num << endl; changeToFive(num); cout << "Now the output is " << num << endl; return 0; } void changeToFive(int num1) { num1 = 5; cout << "The number is now: " << num1 << endl; } ``` #### Analysis 1. **Main Function Execution:** - The program begins by entering the `main()` function. - An integer variable `num` is initialized with the value `1`. - It prints: `"In the main function, the number is: 1"` 2. **Function Call `changeToFive(num)`:** - The value of `num` (which is `1`) is passed to the `changeToFive` function. - Inside `changeToFive`, the variable `num1` is a copy of `num`. - The line `num1 = 5;` changes the local copy `num1` to `5`. - It prints: `"The number is now: 5"` - Note: This change does not affect `num` in the `main()` function. 3. **After the Function Call:** - Control returns to `main()`. - It prints: `"Now the output is 1"` #### Conclusion The output after executing this code will be: ``` In the main function, the number is: 1 The number is now: 5 Now the output is 1 ``` This illustrates the concept of pass-by-value in C++, where modifications to function parameters do not affect the original arguments.
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