Question

Using the ListNode structure introduced in this chapter, write a function void printFirst(ListNode *ptr) that prints the value stored in the first node of a list passed to it as parameter. The function should print an error message and terminate the program if the list passed to it is empty.

Short answer

Question: Implement a function called "printFirst" that takes a pointer to a ListNode and prints the value stored in the first node of the list. If the list is empty, print an error message and terminate the program.

Step-by-step solution

ListNode structure

For reference, I will provide the "ListNode" structure that we will work with: ```cpp struct ListNode { int value; ListNode *next; }; ```

Check if the list is empty

First, we need to check if the list is empty by verifying if the pointer (ptr) passed to the function is a nullptr. This means that there is no first element in the list, and we need to handle this case accordingly. ```cpp if (ptr == nullptr) ```

Print error message and terminate program

If the list is empty (ptr is nullptr), print an error message and terminate the program using the "exit()" function: ```cpp if (ptr == nullptr) { std::cerr << "The list is empty. Cannot print the first value." << std::endl; exit(1); } ```

Print the value of the first node

If the list is not empty, meaning the ptr is not nullptr, print the value of the first node in the list: ```cpp std::cout << "The first value in the list is: " << ptr->value << std::endl; ```

Full Code

Here is the full code for the "printFirst" function: ```cpp #include struct ListNode { int value; ListNode *next; }; void printFirst(ListNode *ptr) { if (ptr == nullptr) { std::cerr << "The list is empty. Cannot print the first value." << std::endl; exit(1); } std::cout << "The first value in the list is: " << ptr->value << std::endl; } ``` This function should be used according to the instructions of the exercise.