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What is the output of the following code?
int x = 10;
int* ptr1 = &x;
int** ptr2 = &ptr1;
cout << **ptr2 << endl;
int x = 10;
int* ptr1 = &x;
int** ptr2 = &ptr1;
cout << **ptr2 << endl;
- a)10
- b)0
- c)Compiler error
- d)Runtime error
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
What is the output of the following code?int x = 10;int* ptr1 = &x...
Explanation:
Pointers Initialization:
- int x = 10; // Initializing integer variable x with value 10
- int* ptr1 = &x; // Creating a pointer ptr1 pointing to the address of x
- int** ptr2 = &ptr1; // Creating a pointer ptr2 pointing to the address of ptr1
Output:
- **ptr2 will dereference the pointer twice, first to get the address of ptr1 and then to get the value at that address, which is the value of x
- Therefore, the output will be the value of x, which is 10
Therefore, the output of the code will be:
a) 10
Pointers Initialization:
- int x = 10; // Initializing integer variable x with value 10
- int* ptr1 = &x; // Creating a pointer ptr1 pointing to the address of x
- int** ptr2 = &ptr1; // Creating a pointer ptr2 pointing to the address of ptr1
Output:
- **ptr2 will dereference the pointer twice, first to get the address of ptr1 and then to get the value at that address, which is the value of x
- Therefore, the output will be the value of x, which is 10
Therefore, the output of the code will be:
a) 10
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Community Answer
What is the output of the following code?int x = 10;int* ptr1 = &x...
'ptr2' is a pointer to a pointer. It holds the address of 'ptr1', which in turn holds the address of 'x'. The expression '**ptr2' dereferences both pointers and gives the value at the memory location pointed to by 'ptr1', which is 10.
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