Accessing structure members in C++ - Mini Project: Build & Apply

Accessing structure members

📖 Scenario: You are creating a simple program to store and display information about a book in a library.

🎯 Goal: Build a program that defines a structure for a book, stores information in it, and prints the book's details by accessing its members.

📋 What You'll Learn

Define a structure named Book with members title, author, and year.

Create a variable of type Book and assign values to its members.

Access and print the members of the Book variable.

💡 Why This Matters

🌍 Real World

Structures help organize related data together, like storing details about books, students, or products in real applications.

💼 Career

Understanding how to define and access structures is fundamental for many programming jobs, especially in systems programming, game development, and software engineering.

1

Define the Book structure

Define a structure called Book with three members: title and author as std::string, and year as int.

C++

#include 
#include 

// Define the Book structure here

int main() {

    return 0;
}

Hint

Use the struct keyword to define Book with the specified members.

2

Create a Book variable and assign values

Inside main(), create a variable named myBook of type Book. Assign "The Great Gatsby" to myBook.title, "F. Scott Fitzgerald" to myBook.author, and 1925 to myBook.year.

C++

#include 
#include 

struct Book {
    std::string title;
    std::string author;
    int year;
};

int main() {
    // Create myBook and assign values

    return 0;
}

Hint

Create a variable of type Book and use the dot . operator to assign values to its members.

3

Access and print the Book members

Use std::cout to print the book details in this exact format:
Title: The Great Gatsby
Author: F. Scott Fitzgerald
Year: 1925
Access the members of myBook using the dot operator.

C++

#include 
#include 

struct Book {
    std::string title;
    std::string author;
    int year;
};

int main() {
    Book myBook;
    myBook.title = "The Great Gatsby";
    myBook.author = "F. Scott Fitzgerald";
    myBook.year = 1925;

    // Print book details here

    return 0;
}

Hint

Use std::cout and the dot operator to print each member on its own line.

4

Run the program to display the book details

Run the program and ensure it prints exactly:
Title: The Great Gatsby
Author: F. Scott Fitzgerald
Year: 1925

C++

#include 
#include 

struct Book {
    std::string title;
    std::string author;
    int year;
};

int main() {
    Book myBook;
    myBook.title = "The Great Gatsby";
    myBook.author = "F. Scott Fitzgerald";
    myBook.year = 1925;

    std::cout << "Title: " << myBook.title << std::endl;
    std::cout << "Author: " << myBook.author << std::endl;
    std::cout << "Year: " << myBook.year << std::endl;

    // Run the program to see output
}

Hint

Make sure your output matches exactly, including capitalization and spacing.

Practice

1. What operator is used to access members of a structure variable in C++?

easy

A. Double colon (::)

B. Arrow operator (->)

C. Colon operator (:)

D. Dot operator (.)

Solution

1. Step 1: Recall structure member access syntax

   In C++, to access a member of a structure variable, the dot operator (.) is used.

2. Step 2: Differentiate from pointer access

   The arrow operator (->) is used when accessing members through a pointer, not a direct variable.

3. Final Answer:

   Dot operator (.) -> Option D

4. Quick Check:

   Structure member access = dot operator (.) [OK]

Hint: Use dot (.) for variables, arrow (->) for pointers [OK]

Common Mistakes:

• Confusing dot (.) with arrow (->)
• Using colon (::) which is for scope resolution
• Using arrow (->) on non-pointer variables

2. Which of the following is the correct syntax to access member age of a structure variable person?

easy

A. person->age;

B. person::age;

C. person.age;

D. person:age;

Solution

1. Step 1: Identify the variable type

   Since person is a structure variable (not a pointer), use the dot operator to access members.

2. Step 2: Check syntax correctness

   Correct syntax is person.age;. The arrow operator (->) is for pointers, colons are invalid here.

3. Final Answer:

   person.age; -> Option C

4. Quick Check:

   Direct struct variable uses dot (.) [OK]

Hint: Use dot (.) for struct variables, arrow (->) for pointers [OK]

Common Mistakes:

• Using arrow (->) on non-pointer variables
• Using colon (:) or double colon (::) incorrectly
• Missing semicolon at end

3. What will be the output of the following code?

#include <iostream>
struct Point {
  int x;
  int y;
};

int main() {
  Point p = {3, 4};
  std::cout << p.x + p.y << std::endl;
  return 0;
}

medium

A. 7

B. 34

C. 0

D. Compilation error

Solution

1. Step 1: Understand structure initialization

   The structure Point is initialized with x=3 and y=4.

2. Step 2: Calculate the output expression

   The code outputs p.x + p.y which is 3 + 4 = 7.

3. Final Answer:

   7 -> Option A

4. Quick Check:

   3 + 4 = 7 [OK]

Hint: Add members accessed by dot operator for result [OK]

Common Mistakes:

• Concatenating numbers as strings (34)
• Assuming default zero initialization
• Confusing pointer access syntax

4. Identify the error in the following code snippet:

struct Car {
  int speed;
};

int main() {
  Car c;
  c->speed = 100;
  return 0;
}

medium

A. Missing semicolon after struct definition

B. Using arrow operator (->) on a non-pointer variable

C. Incorrect struct member name

D. No error, code is correct

Solution

1. Step 1: Check variable type and operator

   The variable c is a struct variable, not a pointer.

2. Step 2: Identify correct member access

   Using c->speed is invalid here; should use c.speed.

3. Final Answer:

   Using arrow operator (->) on a non-pointer variable -> Option B

4. Quick Check:

   Non-pointer struct uses dot (.) [OK]

Hint: Use dot (.) for variables, arrow (->) only for pointers [OK]

Common Mistakes:

• Using arrow (->) on struct variables
• Assuming arrow works for all struct access
• Ignoring pointer vs variable difference

5. Given the following code, how can you correctly access the member name of the structure pointer ptr?

struct Student {
  std::string name;
  int id;
};

Student s = {"Alice", 101};
Student* ptr = &s;
// Access name here

hard

A. ptr->name;

B. ptr.name;

C. ptr::name;

D. *ptr.name;

Solution

1. Step 1: Understand pointer member access

   When accessing members through a pointer, you can use the arrow operator (->) or dereference then dot ( (*ptr).member ).

2. Step 2: Check each option

   ptr.name; is invalid syntax for a pointer. ptr->name; is the correct syntax using the arrow operator. ptr::name; incorrectly uses scope resolution operator. *ptr.name; is wrong due to operator precedence; it should be (*ptr).name;.

3. Final Answer:

   ptr->name; -> Option A

4. Quick Check:

   Pointer member access = arrow (->) or (*ptr).member [OK]

Hint: Use arrow (->) or dereference then dot ( (*ptr).member ) [OK]

Common Mistakes:

• Using dot (.) directly on pointer
• Forgetting to dereference pointer
• Confusing pointer and variable access