Question

Suppose the vector v contains the letters 'A', 'B', 'C', and 'D' in that order. What is the output of the following code? vector: : iterator \(i=v .\) begin () \(i++\) cout \(<<*(i+2)<<"\) "; \(i--\) cout \(<

Short answer

```cpp vector::iterator i = v.begin(); i++; cout << *(i+2) << " "; i--; cout << i[2]; cout << *(i+2) << " "; ``` Answer: The final output of the code snippet is "D C C".

Step-by-step solution

Understand the vector representation

The vector v contains the letters 'A', 'B', 'C', and 'D' in that order. So, we can represent the vector as: v = ['A', 'B', 'C', 'D']

Initialize the iterator

The code initializes an iterator i as follows: vector::iterator i = v.begin(); Initially, the iterator i points to the first element of the vector, i.e., 'A'.

Performing iterator operations

Now, let's perform the iterator operations given in the code. First, increment the iterator i by one step: i++; Now, i points to the second element of the vector, i.e., 'B'.

Output the element using '*(i+2)'

The code now calls for outputting the element two steps ahead of the iterator: cout << *(i+2) << " "; To find which element it refers to, we add 2 steps to the iterator i. Since i is currently pointing at 'B', adding two steps gives us the element 'D'. So, the output will be: D

Decrement the iterator 'i'

Next, the code decrements the iterator i by one step: i--; Now, i points back to the first element of the vector, which is 'A'.

Output the element using 'i[2]'

The code now calls for outputting the element at the index 2 relative to the iterator 'i': cout << i[2]; Since i is pointing to 'A', index 2 relative to 'i' corresponds to the third element of the vector, i.e., 'C'. So, the output will be: C

Output the element using '*(i+2)' again

Finally, the code calls for outputting the element two steps ahead of 'i' once more: cout << *(i+2) << " "; As i is pointing to 'A', two steps ahead corresponds to the element 'C'. Thus, the output will be: C

Combining the outputs

Now that we have the individual outputs for each line of code, let's combine them to get the final output. The combined output is: D C C