Question

include #include #include using namespace std; void traceMe(double x, double y); int main() { double one, t… # Consider the following program: #include #include #include using namespace std; void traceMe(double x, double y); int main() { double one, two; cout << "Enter two numbers: "; cin >> one >> two; cout << endl; traceMe(one, two); traceMe(two, one); return 0; } void traceMe(double x, double y) { double z; if (x != 0) z = sqrt(y) / x; else { cout << "Enter a nonzero number: "; cin >> x; cout << endl; z = floor(pow(y, x)); } cout << fixed << showpoint << setprecision(2); cout << x << ", " << y << ", " << z << endl; } a. What is the output if the input is 3 625? b. What is the output if the input is 24 1024? c. What is the output if the input is 0 196?

Short answer

a: 3.00, 625.00, 8.33; 625.00, 3.00, 0.00. b: 24.00, 1024.00, 1.33; 1024.00, 24.00, 0.00. c: 2.00, 196.00, 38416.00; 196.00, 0.00, 0.00.

Step-by-step solution

Understanding the Program Structure

This program consists of a main function and a traceMe function. The main function accepts two numbers from the user as inputs. It then calls the traceMe function twice: first with 'one' as x and 'two' as y, and second with 'two' as x and 'one' as y.

Analyzing Function traceMe

The traceMe function calculates and prints a value 'z' based on the inputs x and y. If x is not zero, it calculates z as the square root of y divided by x. If x is zero, it asks the user to input a new value for x and calculates z as the floor of y raised to the power of x. The output shows x, y, and z formatted to two decimal places.

Solving Part a - Input: 3 625

First call traceMe(3, 625): Since 3 is not zero, calculate z = sqrt(625) / 3 = 25 / 3 = 8.33. Output: 3.00, 625.00, 8.33. Second call traceMe(625, 3): Since 625 is not zero, calculate z = sqrt(3) / 625 = 0.0548 / 625 = 0.0 (after formatting). Output: 625.00, 3.00, 0.00.

Solving Part b - Input: 24 1024

First call traceMe(24, 1024): Since 24 is not zero, calculate z = sqrt(1024) / 24 = 32 / 24 = 1.33. Output: 24.00, 1024.00, 1.33. Second call traceMe(1024, 24): Since 1024 is not zero, calculate z = sqrt(24) / 1024 = 0.1549 / 1024 = 0.0 (after formatting). Output: 1024.00, 24.00, 0.00.

Solving Part c - Input: 0 196

First call traceMe(0, 196): Since x is zero, the program will prompt for a new nonzero input, e.g., x = 2. After receiving the input, calculate z = floor(196 ^ 2) = 38416. Output: 2.00, 196, 38416.00. Second call traceMe(196, 0): Since 196 is not zero, calculate z = sqrt(0) / 196 = 0/196 = 0.0. Output: 196.00, 0.00, 0.00.

Key concepts

Input/Output Operations

In C++, input and output operations allow programs to interact with users or other systems by receiving inputs and displaying outputs. These operations are crucial for gathering necessary data for processing and providing results back to the user. Two primary libraries that facilitate these operations are `` and ``.

The `` library includes utilities for standard input and output operations, namely `cin` and `cout`. The `cin` stream is used to take inputs from the user, while the `cout` stream outputs information. In the given exercise, when the program prompts "Enter two numbers:", it utilizes `cout` to display the prompt and `cin` to store the user's input into variables `one` and `two`. These inputs are then used later in the program to perform calculations.

The `` library, on the other hand, is used to format the output. For instance, the command `cout << fixed << showpoint << setprecision(2);` ensures that the output displays numbers with two decimal places, which is crucial for readability and precision in outputs. The `fixed` manipulator ensures fixed-point notation, while `showpoint` ensures a decimal point is displayed for floating-point numbers.

Control Structures

Control structures in C++ direct the flow of execution within a program. They determine how and when certain parts of a program are executed based on specific conditions.

In the exercise, the `traceMe` function utilizes an `if` statement to decide how to calculate the variable `z`. The `if` statement checks whether the input value `x` is zero. If `x` is not zero, the program calculates `z` as the square root of `y` divided by `x`; otherwise, it requests a new non-zero input for `x`. Therefore, `if` statements are crucial for branching decision points based on conditions within the code.

Another control structure inherently used is the function call itself, where the `main` function calls `traceMe` with different arguments. Each call represents a sequence in which a block of code (the function) is executed when called, further demonstrating how control structures manage flow among different parts of the code.

Data Types

Data types in C++ are essential as they define what types of data can be stored in a variable and what operations can be performed on them. In the exercise program, several data types are utilized, with `double` being the primary one.

The `double` data type is used to store floating-point numbers that can have fractional parts, offering considerable precision necessary for mathematical operations. Variables `one`, `two`, and `z`, all of the `double` type, handle calculations involving division and square roots. This choice is crucial for precision, particularly for situations where calculations yield non-integer results.

Another implicit data type encountered in the exercise is `int`, which is used in operations where integer values are either intended or emerge as results, such as in the `floor` function. The `floor` function converts a floating-point result into its largest integer value not greater than the input value, and using it with `double` inputs produces clean integer outputs stored in `z` after being cast implicitly. Understanding and choosing appropriate data types significantly impact a program’s accuracy, performance, and memory usage.