The given code snippet is a C++ program that demonstrates the use of pointers and functions. Here is the transcription of the code and its analysis:---Assume that the following code segment successfully runs and the address of x is 0x61fe88. What will be displayed on the console?```cppint main(){ double x = 10; double *p = &x; fun(p); cout << x << endl; cout << p << endl; return 0;}void fun(double *&q){ (*q) /= 2; q++;}```### Code Explanation1. **Variable Declaration and Initialization**: - `double x = 10;` initializes a double variable `x` with a value of 10. - `double *p = &x;` declares a pointer `p` that stores the address of `x`.2. **Function Call**: - `fun(p);` calls the function `fun` and passes the pointer `p` by reference.3. **Function Definition**: - `void fun(double *&q)` is a function that takes a reference to a double pointer. - Inside the function, `(*q) /= 2;` halves the value pointed to by `q`. Since `q` points to `x`, the value of `x` becomes 5. - `q++;` increments the pointer `q`, moving it to the next memory address.4. **Console Output**: - `cout << x << endl;` outputs the value of `x`, which is now 5. - `cout << p << endl;` outputs the address stored in `p`. Since `p` was incremented, it no longer points to the original address of `x` (0x61fe88) and now points to a subsequent memory location.### Expected Console Output- The value `5` will be displayed for `x`.- The address printed will be one double memory size ahead of `0x61fe88`.Note: The exact second address output is platform-dependent and may vary depending on system architecture.---

Introduction: The main function has two variables of the double types such as x and p. The variable…

Assume that the following code segment successfully runs and the address of x is Ox61fe88. What will be displayed on the console? int main() { double x = 10; double *p = &x; %3D fun(p); cout << x <« endl; cout << p <« endl; return 0; void fun(double *&q) (*q) /= 2; ++b

Assume that the following code segment successfully runs and the address of x is Ox61fe88. What will be displayed on the console? int main() { double x = 10; double p = &x; %3D fun(p); cout << x <« endl; cout << p <« endl; return 0; void fun(double &q) (*q) /= 2; ++b

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

Problem 1PE

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Concept explainers

Declaring and Defining the function

A set of statements, which performs a particular task is called a function. The program gains modularity by using function. The application is quite easy to understand, read, and edit when they are created with functions. The developer/programmer creates a function and uses main() to call the function. Once the function gets executed, the control is transferred to the main program.

Question

The given code snippet is a C++ program that demonstrates the use of pointers and functions. Here is the transcription of the code and its analysis:

---

Assume that the following code segment successfully runs and the address of x is 0x61fe88. What will be displayed on the console?

```cpp
int main()
{
double x = 10;
double *p = &x;

fun(p);
cout << x << endl;
cout << p << endl;
return 0;
}

void fun(double *&q)
{
(*q) /= 2;
q++;
}
```

### Code Explanation

1. **Variable Declaration and Initialization**:
- `double x = 10;` initializes a double variable `x` with a value of 10.
- `double *p = &x;` declares a pointer `p` that stores the address of `x`.

2. **Function Call**:
- `fun(p);` calls the function `fun` and passes the pointer `p` by reference.

3. **Function Definition**:
- `void fun(double *&q)` is a function that takes a reference to a double pointer.
- Inside the function, `(*q) /= 2;` halves the value pointed to by `q`. Since `q` points to `x`, the value of `x` becomes 5.
- `q++;` increments the pointer `q`, moving it to the next memory address.

4. **Console Output**:
- `cout << x << endl;` outputs the value of `x`, which is now 5.
- `cout << p << endl;` outputs the address stored in `p`. Since `p` was incremented, it no longer points to the original address of `x` (0x61fe88) and now points to a subsequent memory location.

### Expected Console Output

- The value `5` will be displayed for `x`.
- The address printed will be one double memory size ahead of `0x61fe88`.

Note: The exact second address output is platform-dependent and may vary depending on system architecture.

---

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