What is the output of the following code fragment? int $$ \mathrm{v} 1=2, \mathrm{v} 2=-1,{ }^{*} \mathrm{p} 1,{ }^{*} \mathrm{p} 2 ; $$ $$ \mathrm{p} 1=\& \mathrm{v} 1 ; $$ $$ \mathrm{p} 2=\& \mathrm{v} 2 ; $$ $$ \mathrm{p} 2=\mathrm{p} 1 ; $$ cout

Question

What is the output of the following code fragment? int $$ \mathrm{v} 1=2, \mathrm{v} 2=-1,{ }^{*} \mathrm{p} 1,{ }^{*} \mathrm{p} 2 ; $$ $$ \mathrm{p} 1=\& \mathrm{v} 1 ; $$ $$ \mathrm{p} 2=\& \mathrm{v} 2 ; $$ $$ \mathrm{p} 2=\mathrm{p} 1 ; $$ cout << $$ { }^{*} \mathrm{p} 2 \ll $$ endl; $$ \frac{1}{-2} $$ -1

UpStudy AI · Accepted Answer

Let's analyze the given C++ code step by step to determine its output.

### Code Fragment:
```cpp
#include 
using namespace std;

int main() {
    int v1 = 2, v2 = -1, *p1, *p2;
    p1 = &v1;
    p2 = &v2;
    p2 = p1;
    cout << *p2 << endl;
    return 0;
}
```

### Step-by-Step Execution:

1. **Variable Initialization:**
   ```cpp
   int v1 = 2, v2 = -1, *p1, *p2;
   ```
   - `v1` is initialized to `2`.
   - `v2` is initialized to `-1`.
   - `p1` and `p2` are pointers to integers, but they are not yet assigned any address.

2. **Assigning Addresses to Pointers:**
   ```cpp
   p1 = &v1;
   p2 = &v2;
   ```
   - `p1` now holds the address of `v1`.
   - `p2` now holds the address of `v2`.

3. **Reassigning `p2` to Point to `v1`:**
   ```cpp
   p2 = p1;
   ```
   - Now, `p2` points to the same location as `p1`, which is the address of `v1`.

4. **Dereferencing `p2` and Printing:**
   ```cpp
   cout << *p2 << endl;
   ```
   - Since `p2` now points to `v1`, `*p2` gives the value of `v1`, which is `2`.
   - Therefore, `2` is printed to the console.

### Conclusion:
After executing the code, the output will be:

```
2
```

**Note:** The options you provided (`1/(-2)` and `-1`) do not align with the actual output of the code. The correct output is `2`.
The code prints `2` to the console.

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