Question

Consider the following statements: linkedStackType stack; linkedQueueType queue; int num; Suppose the input is: 48 35 72 88 92 11 10 15 44 52 67 36 Show what is written by the following segment of code: stack.push(0); queue.addQueue(0); cin >> num; while (cin) { switch (num % 3) { case 0: stack.push(num); break; case 1: queue.addQueue(num); break; case 2: if (!stack.isEmptyStack()) { cout << stack.top() << " "; stack.pop(); } else if (!queue.isEmptyQueue()) { cout << queue.front() << " "; queue.deleteQueue(); } else cout << "Stack and queue are empty." << endl; break; } //end switch cin >> num; } //end while cout << endl; cout << "stack: "; while (!stack.isEmptyStack()) { cout << stack.top() << " "; stack.pop(); } cout << endl; cout << "queue: "; while (!queue.isEmptyQueue()) { cout << queue.front() << " "; queue.deleteQueue(); } cout << endl;

Short answer

The output is `72 72 48 0`, followed by `stack: 48` and `queue: 0 35 88 92 67`. The code prints stack and queue contents post-processing.

Step-by-step solution

Initialize and Input

The initial step involves setting up the stack and queue, and then reading the first number from the input. The stack initially has a single element: `0`. Similarly, the queue also contains the single element: `0`. The first input number, `num`, is read as `48`.

Process Each Number

We proceed to evaluate each number via a `while(cin)` loop and a `switch` statement based on `num % 3`: - For `48` and `72` (both `num % 3 == 0`), push these values onto the `stack`. - For `35`, `88`, `92`, and `67` (all `num % 3 == 1`), add these values to the `queue`. - For `11`, `10`, `15`, `44`, `36` (all `num % 3 == 2`), perform the action of printing and removing the `top` of the `stack`. If `stack` is empty, remove from and print the front of the `queue` instead. After processing all numbers, the output generated from the conditions in the `switch` statement is: `72 72 48 0`.

Post-Processing - Display Stack

Once all input numbers have been processed, we display the remaining contents of the `stack`. The stack is processed by continuously popping the top and printing it until the stack is empty. The stack currently contains only the number `48`, following the last operation where it added `48` during the case 0 condition and was not removed subsequently.

Post-Processing - Display Queue

Finally, the queue's contents are shown by removing the front element and printing it until the queue is empty. The queue contains the numbers `0`, `35`, `88`, `92`, `67`, after all operations are complete. These numbers are displayed in the order they were inserted.

Key concepts

Stack

A stack is a linear data structure known for its Last In First Out (LIFO) property. This means that the last element added to the stack will be the first to be removed.
It is similar to piling up plates; you can only remove the top plate first.
In programming and algorithms, stacks are used in various operations such as parsing expressions, managing function calls, and backtracking problems. In the provided exercise, the stack starts with a single element `0`. As input numbers are processed, numbers like `48` and `72` that satisfy `num % 3 == 0` are pushed onto the stack.

• When a number qualifies to be pushed onto the stack, it means it falls into the case where its remainder when divided by 3 is `0`.
• If `num % 3 == 2` and the stack is not empty, the top number is printed and then removed.

Hence, at the end of the operations and output printing, only the elements that couldn't be popped remain in the stack.

Queue

A queue is also a linear data structure but follows the First In First Out (FIFO) principle. Imagine a line of people at a service counter; the person at the front of the line gets served first.
This structure is used in scheduling processes in computing, navigating breadth-first searches in graphs, and handling requests in a controlled sequence. In the given exercise, the queue also begins with just `0`. As input is provided, numbers satisfying `num % 3 == 1` such as `35`, `88`, `92`, and `67` are added to the queue.

• The queue sees usage in the case where the stack may be empty, providing a fallback means to maintain operation continuity by dequeuing and outputting its front element.
• This mechanism allows the queue to manage and output elements that match a different set of conditions compared to the stack.

Ultimately, numbers like `0`, `35`, `88`, `92`, `67` remain post-processing due to their insertion order and any popping dynamics that occur.

C++ Programming

C++ is a robust programming language commonly used for system/software development, game programming, and real-time processing. It is known for its performance and use in writing efficient algorithms and data structure handling. The exercise in question utilizes C++ to model and manipulate stack and queue data structures, demonstrating how versatile and efficient C++ can be in handling complex data operations. In C++, the implementation of the stack and queue as illustrated usually involves the use of library header files like `#include` and `#include`, or custom implementations, as seen with `linkedStackType` and `linkedQueueType`.
The code snippet also demonstrates the use of input handling (`cin`), decision-making with `switch` cases, and looping constructs to steer program flow—key parts of C++ programming mastery.

Algorithms

An algorithm is essentially a step-by-step procedure to perform a task or solve a problem. In computer science, algorithms form the backbone of task execution and data management. For the provided exercise, an algorithm decides how numbers are processed and outputted based on their value modulo 3. This provides a deterministic system where:

• `num % 3 == 0` adds numbers to a stack;
• `num % 3 == 1` queues numbers, and;
• `num % 3 == 2` triggers output from the stack (or queue if the stack is empty).

Such algorithms demonstrate basic control flow and provide building blocks for more complex operations like error handling, data parsing, and simulation processes.
By understanding algorithm dynamics here, one prepares for crafting efficient and effective solutions to broader programming challenges.