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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Transcribed Image Text: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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