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Consider the following type definition: struct ShoeType \\{ char style double price \\}; Given this structure type definition, what will be the output produced by the following code? ShoeType shoe1, shoe2; shoe1.style \(=^{\prime} A^{\prime}\) shoe1.price \(=9.99\) cout \( < < \) shoe \(1 .\) style \( < < " \$^{\prime \prime} < < \) shoe1.price \( < < \) endl shoe \(2=\) shoe 1 shoe2.price \(=\) shoe \(2 .\) price \(/ 9\) cout \( < < \) shoe \(2 .\) style \( < < " \$ " < < \) shoe \(2 .\) price \( < < \) end 1;

Short Answer

Expert verified
Answer: The printed output values are "A $9.99" for shoe1 and "A $1.11" for shoe2.

Step by step solution

01

Analyze the structure definition

struct ShoeType { char style; double price; }; The struct ShoeType has two members - style (a character) and price (a double).
02

Initialize shoe1 and shoe2 objects

ShoeType shoe1, shoe2; Here, two objects of type ShoeType, shoe1 and shoe2, are created.
03

Assign values to shoe1

shoe1.style = 'A'; shoe1.price = 9.99; The character 'A' is assigned to the style member of shoe1, and the value 9.99 is assigned to the price member of shoe1.
04

Print shoe1 values

cout << shoe1.style << " $" << shoe1.price << endl; The style and price of shoe1 are printed, separated by a " \(". The output will be: A \)9.99
05

Assign shoe1 values to shoe2

shoe2 = shoe1; Both the style and price members of shoe2 are assigned the corresponding values of shoe1, which means shoe2.style equals 'A' and shoe2.price equals 9.99.
06

Update shoe2 price

shoe2.price = shoe2.price / 9; The price of shoe2 is divided by 9 and assigned back to shoe2.price. So, shoe2.price = 9.99 / 9 \(\approx 1.11\).
07

Print shoe2 values

cout << shoe2.style << " $" << shoe2.price << endl; The style and updated price of shoe2 are printed, separated by a " \(". The output will be: A \)1.11 To summarize, the output of the given code is: A $9.99 A $1.11

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Data Structures
In C++, data structures are essential tools for organizing and storing data efficiently. They allow you to create a custom blueprint to encapsulate different kinds of information under one name. This abstraction is crucial when handling complex data in sophisticated applications.

Consider them like real-world blueprints for objects such as buildings or vehicles, which detail various components and how they relate to each other. In programming terms, data structures allow you to organize data logically, making it easy to perform operations on it.
  • Arrays: Store a collection of elements, all of the same type.
  • Linked Lists: A sequence of nodes where each node points to the next one, allowing for dynamic memory allocation.
  • Stacks: Follows a Last In, First Out (LIFO) approach.
  • Queues: Follows a First In, First Out (FIFO) approach.
  • Structs: A composite data structure that groups variables of different types under a single name.
Each type has specific use-cases based on the needs of your application, with structs being particularly useful in C++ for grouping various data under one unified entity.
Structs in C++
Structs in C++ are composite data types that provide a way to group variables of different data types together under a single name. They are a fundamental part of the language, allowing you to model real-world entities by combining various data types.

Structs are declared using the `struct` keyword followed by the structure's name and its variables enclosed in curly braces. Here's the syntax for defining a struct:

`struct StructName {`
`// variable declarations`
`};`

Structs are particularly handy because they offer a simplistic way to manage related data. For example, in a struct defining a `ShoeType`, you can encapsulate both the style and the price in one entity. This keeps your code organized and makes it easy to develop complex data models.

Another key benefit of structs is their similarity to classes in C++, but with default public access for their members. They are especially useful when you need to group data without the overhead of complex operations, as would be the case with a class.
Object Initialization
Initialization is a fundamental concept in dealing with objects in C++. When you initialize an object, you are assigning initial values to its data members. This sets up your object to function correctly from the start.

In the example provided, `ShoeType shoe1, shoe2;` initializes two objects of type ShoeType. Each of these objects can separately encapsulate different styles and prices. Using a structured form of initialization can significantly reduce errors and make your code more logical and readable.

Initialization can also ensure that default values are in place and help prevent undefined behavior, which can occur when you attempt to use uninitialized data. Be it for built-in types or user-defined types, initialization cannot be overlooked. An initialized object is like a prepared workspace, ready for operations without the risk of unexpected errors.
Assignment in C++
Assignment in C++ is the process of giving a value to a variable or object. It is a core part of programming, allowing you to manipulate the data within your structures effectively.

The `=` operator is used for assignment, transferring the right-hand value into the left-hand variable or object member. For instance, `shoe1.style = 'A';` assigns the character 'A' to the `style` member of `shoe1`. Similarly, assignments can work with entire objects, like `shoe2 = shoe1;`, which copies all the data from `shoe1` into `shoe2`.

This operation can overwrite existing values or, in the case of compounding assignments, like `shoe2.price = shoe2.price / 9;`, modify the current value using arithmetic operations.

Assignments are fundamental as they allow changes in your program's state. They are essential in dynamic applications where data is expected to change according to input or other conditions.

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Most popular questions from this chapter

Give a definition for the function with the following function declaration. The class BankAccount is defined in Display 10.5 BankAccount new_account(BankAccount old_account); //Precondition: old_account has previously been given a value // (that is, its member variables have been given values). //Returns the value for a new account that has a balance of zero / / and the same interest rate as the old_account. For example, after this function is defined, a program could contain the following: BankAccount account3, account4; account3.set(999, 99, 5.5) ; account4 \(=\) new_account(account 3) account4.output(cout); This would produce the following output:

Given the following struct definition: struct \(A\) \\{ int member \(_{-} b\) int member_c; \\}; declare \(x\) to have this structure type. Initialize the members of \(x\) member \(_{-} b\) and member_c, to the values 1 and 2 , respectively. Note: This requests an initialization, not an assignment of the members. This distinction is important and will be made in the text in a later chapter.

Give a function definition corresponding to the following function declaration. (The type ShoeType is given in Self-Test Exercise \(2 .)\) ShoeType discount(ShoeType old_record); //Returns a structure that is the same as its argument, //but with the price reduced by \(10 \%\).

Here is an initialization of a structure type. Tell what happens with each initialization. Note any problems with these initializations. struct Date \\{ int month; int day; int year; \\}; a. Date due_date \(=\\{12,21\\}\) b. Date due_date \(=\\{12,21,2022\\}\) c. Date due_date \(=\\{12,21,20,22\\}\) d. Date due_date \(=\\{12,21,22\\}\)

Write a definition for a structure type for records consisting of a person's wage rate, accrued vacation (which is some whole number of days), and status (which is either hourly or salaried). Represent the status as one of the two char values 'H' and 'S'. Call the type EmployeeRecord.

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