Question

Consider the following statements: stackType stack; int x; Suppose that the input is: 14 45 34 23 10 5 -999 Show what is output by the following segment of code: stack.push(5); cin >> x; while (x != -999) { if (x % 2 == 0) { if (!stack.isFullStack()) stack.push(x); } else cout << "x = " << x << endl; cin >> x; } cout << "Stack Elements: "; while (!stack.isEmptyStack()) { cout << " " << stack.top(); stack.pop(); } cout << endl;

Short answer

Odd numbers shown as output are: 45, 23, 5. Final stack output is: 10 34 14 5.

Step-by-step solution

Initialize Variables and Input

The program initializes an integer stack and a variable `x`. It then processes the input values in sequence. The input provided is `14 45 34 23 10 5 -999`.

Push Initial Value to Stack

Before processing the input list, the stack receives the integer `5` by executing `stack.push(5);`. Therefore, the stack initially contains: `5`.

Begin Input Loop and Process Even Numbers

The loop starts by reading the first input `14`. Since `14` is even (`14 % 2 == 0`), it's checked if the stack is full. Assuming the stack isn't full, `14` is pushed to the stack. The stack now contains: `5, 14`.

Handle Odd Number: Output and Continue

Next, `45` (odd number) is read. Hence, it is printed as `x = 45`. The stack remains unchanged: `5, 14`.

Continue Even Number Processing

The next input `34` is even, so it's pushed onto the stack. The stack now contains: `5, 14, 34`.

Handling Another Odd Number

The next read value is `23`, which is odd. This value outputs as `x = 23`. The stack remains: `5, 14, 34`.

Process Another Even and Odd Number

The value `10` is read next, which is even, thus it's pushed onto the stack. The stack now updates to: `5, 14, 34, 10`. The next value `5` is odd and outputs as `x = 5`. The stack remains unchanged: `5, 14, 34, 10`.

Output Stack Elements

The input sequence ends at `-999`, terminating the loop. The stack contents are printed by popping each element: `10, 34, 14, 5`.

Key concepts

Problem Analysis

Understanding the problem is crucial before diving into the code. Here, we have a stack data structure implemented using a fictional `stackType` class in C++. The task involves reading a sequence of integers, processing them based on whether they are even or odd, and subsequently printing out the stack's content. The stopping condition, indicated by `-999`, signals when to end processing.

The main operations performed on the stack are `push`, which adds an element, and `pop`, which removes the top element. Additionally, we check if the stack is full with `isFullStack()` and if it's empty with `isEmptyStack()`. Only even numbers are pushed onto the stack, while odd numbers are directly output. This problem involves careful control and understanding of loops, conditions, and stack operations.

Even and Odd Number Handling

In this exercise, the distinction between even and odd numbers dictates the program flow. Even numbers are those divisible by 2, meaning when you compute `x % 2`, it results in 0. Odd numbers, on the other hand, produce a remainder of 1 when divided by 2.

The code uses this property to decide how to handle the input numbers:

• If the number is even (e.g., 14, 34, 10), it is pushed onto the stack, but only if the stack isn't full.
• Odd numbers, like 45, 23, and another 5, are printed directly and are not pushed onto the stack.

Through this simple condition checking, the code processes inputs efficiently while adhering to the exercise's requirements.

Input Loop Processing

The input loop plays a crucial role in handling multiple numbers sequentially. This loop continues to run as long as the input value is not `-999`. This signifies the end of input processing and ensures values are correctly managed until the stop value is encountered.

The `while` loop structure here is set up as follows:

• Read an input using `cin >> x`.
• Check if `x` is not equal to `-999`. If so, process it. Otherwise, exit the loop.

The order of these operations ensures every number is accounted for and processed just once. It handles each number based on its evenness or oddness before reading the next input.

Stack Implementation

Stack implementation in C++ is facilitated through collections of stack operations provided by `stackType`. This exercise assumes `stackType` contains methods like `push(int x)`, `pop()`, `top()`, `isFullStack()`, and `isEmptyStack()`. These methods manage the stack's contents from initialization through manipulation and eventual output.

Key stack operations used here:

• `push(x)`: Adds the element `x` to the top of the stack.
• `pop()`: Removes the topmost element from the stack.
• `top()`: Retrieves the topmost element without removing it.
• `isFullStack()`: Checks if the stack can accommodate more elements.
• `isEmptyStack()`: Verifies if the stack is depleted of elements.

Implementing these operations ensures efficient management and correct ordering of elements, a pivotal aspect of stack-based algorithms.

Output Formatting

The final part of this problem involves correctly formatting and displaying the output. After processing all inputs, the stack's content is printed. This requires using a loop to continuously pop elements from the stack until it's empty.

The output process involves:

• Printing "Stack Elements:" as the initial statement.
• Popping each element with `stack.pop()` and printing it. This retrieves elements in reverse order of their insertion, demonstrating the Last-In, First-Out (LIFO) property of stacks.
• Each popped element is printed along with a space for formatting considerations.

This approach highlights the contents effectively and presents them in a clean, readable format, demonstrating the stack's sequential data arrangement.