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
Ask by Mcfarlane Chandler. in the United States
Jan 21,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The code prints `2` to the console.
Solution
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`.
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Bonus Knowledge
When you run this code fragment, the output will be `2`. Here's the scoop: initially, `p1` points to `v1` which has a value of `2`, and `p2` points to `v2` which holds `-1`. However, when you do `p2 = p1;`, `p2` now points to `v1`. So, when you dereference `p2` with `*p2`, you're actually getting the value of `v1`, which is `2`. Then, the `cout` prints this value followed by a new line. If you're curious about pointer operations, remember that pointers are like signs pointing to houses - change the sign's direction, and you're knocking on a different door! So, always keep an eye on where your pointers are pointing!
