Question

Predict what the output will be. ListNode *p \(=\) new ListNode (56.4); \(p=\) new ListNode \((34.2, p):\) ListNode *q \(=p->\) next cout \(\langle\langle q\text { -) value }\)

Short answer

Answer: 56.4

Step-by-step solution

Understand the ListNode class

To understand the given code, we first need to know the ListNode class's structure. ListNode class has two members: a floating-point number 'value' and a pointer to the next ListNode object 'next'. The value of the ListNode is stored in 'value', while the pointer to the next ListNode is stored in 'next'.

Create a new ListNode object p

In this step, a new ListNode object 'p' with the value 56.4 is created. The syntax for this operation is: ListNode *p = new ListNode (56.4); At this point, the linked list only contains one element, pNode with value 56.4, and the next pointer is set to NULL since there's no next element initialized.

Add a new ListNode object with value 34.2 before p

The code adds another ListNode object with the value 34.2 before the existing ListNode object p. The syntax is: p = new ListNode (34.2, p); At this point, the linked list now contains two elements: pNode with value 34.2 pointing to the pNode with value 56.4.

Create a new ListNode pointer q and set it to point to the next element of p

Now, a new pointer 'q' is created and set to point to the next element of the ListNode object 'p'. This is done by setting q to the value of p->next. The syntax is: ListNode *q = p->next; At this point, the pointer q is pointing to the ListNode object with the value 56.4.

Output the value of ListNode pointed by q

The last step is to output the value of the ListNode that 'q' is pointing to. The syntax for this operation is: cout << q->value; Since q is pointing to the ListNode object with the value 56.4, the output will be 56.4. Final output: 56.4