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What is the output of the following code snippet?
int x = 5;
int* ptr = &x;
int** ptr2 = &ptr;
cout << **ptr2;
int x = 5;
int* ptr = &x;
int** ptr2 = &ptr;
cout << **ptr2;
- a)5
- b)10
- c)The program will not compile
- d)Garbage value
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
What is the output of the following code snippet?int x = 5;int* ptr = ...
The code snippet initializes an integer variable 'x' with the value 5. A pointer variable 'ptr' is declared and assigned the address of 'x'. Then, a pointer to a pointer variable 'ptr2' is declared and assigned the address of 'ptr'. Dereferencing 'ptr2' twice using the asterisk (*) operator ('**ptr2') gives the value at the memory location pointed to by 'ptr', which is 'x'. Therefore, the output is 5.
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Community Answer
What is the output of the following code snippet?int x = 5;int* ptr = ...
Understanding the Code Snippet
The provided code snippet is a simple demonstration of pointers in C++. Let's break down the code to understand its output.
Code Breakdown
- Variable Declaration:
- `int x = 5;` declares an integer variable `x` and initializes it to 5.
- Pointer Initialization:
- `int* ptr = &x;` creates a pointer `ptr` that holds the address of `x`.
- Pointer to Pointer:
- `int ptr2 = &ptr;` creates another pointer `ptr2` that holds the address of the pointer `ptr`.
Dereferencing the Pointer to Pointer
- Dereferencing Explained:
- When we use `ptr2`, we first dereference `ptr2` to get `ptr`, and then we dereference `ptr` to get the value of `x`.
- Execution of `cout < />ptr2;`:
- This will output the value of `x`, which is 5.
Conclusion
- The output of the code is `5`, making option 'A' the correct answer. There are no compilation errors, and the program will run successfully, yielding the expected output.
Key Points
- `x` is initialized to 5.
- `ptr` points to `x`.
- `ptr2` points to `ptr`.
- `ptr2` accesses the value of `x`, which is 5.
This straightforward example demonstrates how pointers and pointer-to-pointer constructs work in C++.
The provided code snippet is a simple demonstration of pointers in C++. Let's break down the code to understand its output.
Code Breakdown
- Variable Declaration:
- `int x = 5;` declares an integer variable `x` and initializes it to 5.
- Pointer Initialization:
- `int* ptr = &x;` creates a pointer `ptr` that holds the address of `x`.
- Pointer to Pointer:
- `int ptr2 = &ptr;` creates another pointer `ptr2` that holds the address of the pointer `ptr`.
Dereferencing the Pointer to Pointer
- Dereferencing Explained:
- When we use `ptr2`, we first dereference `ptr2` to get `ptr`, and then we dereference `ptr` to get the value of `x`.
- Execution of `cout < />ptr2;`:
- This will output the value of `x`, which is 5.
Conclusion
- The output of the code is `5`, making option 'A' the correct answer. There are no compilation errors, and the program will run successfully, yielding the expected output.
Key Points
- `x` is initialized to 5.
- `ptr` points to `x`.
- `ptr2` points to `ptr`.
- `ptr2` accesses the value of `x`, which is 5.
This straightforward example demonstrates how pointers and pointer-to-pointer constructs work in C++.
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