Nested structures in C++ - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a nested structure inside another structure.

C++

struct Outer {
    struct [1] {
        int x;
    } inner;
};

Drag options to blanks, or click blank then click option'

AInner

BNested

Cinner

DInnerStruct

Attempts:

3 left

💡 Hint

Common Mistakes

Using the member variable name instead of the structure name.

Using lowercase for the structure name.

The nested structure inside Outer is named 'Inner'.

2fill in blank

medium

Complete the code to access the member 'x' of the nested structure.

C++

Outer::Inner obj;
obj.[1] = 10;

Drag options to blanks, or click blank then click option'

Az

By

Cinner

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to access a member that does not exist.

Confusing the nested structure name with its member.

The member 'x' is defined inside the nested structure Inner.

3fill in blank

hard

Fix the error in the code to correctly initialize the nested structure member.

C++

struct Outer {
    struct Inner {
        int x;
    } [1];
};

Outer o = {{{{10}}}};

Drag options to blanks, or click blank then click option'

Ainner

BInner

CinnerObj

DobjInner

Attempts:

3 left

💡 Hint

Common Mistakes

Using the structure name instead of the member variable name.

Using a different variable name not declared.

The member variable of type Inner is named 'inner', so it must be used in initialization.

4fill in blank

hard

Fill both blanks to define and initialize a nested structure with two members.

C++

struct Outer {
    struct [1] {
        int a;
        int [2];
    } inner;
};

Drag options to blanks, or click blank then click option'

AInner

Bb

Cc

DInnerStruct

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect member names.

Using different structure names.

The nested structure is named 'Inner' and the second member is 'b'.

5fill in blank

hard

Fill all three blanks to create a nested structure and initialize its members.

C++

struct Outer {
    struct [1] {
        int [2];
        int [3];
    } inner;
};

Drag options to blanks, or click blank then click option'

AInner

Ba

Cb

Dc

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong member names or structure names.

Confusing member names with structure names.

The nested structure is named 'Inner' with members 'a' and 'b'.

Practice

1. What is the main purpose of using nested structures in C++?

easy

A. To allow structures to inherit functions

B. To create multiple inheritance between classes

C. To increase the size of a structure unnecessarily

D. To group related data logically inside another structure

Solution

1. Step 1: Understand nested structures concept

   Nested structures let you put one structure inside another to group related data logically.

2. Step 2: Compare options

   Only To group related data logically inside another structure correctly describes grouping data logically. Other options describe unrelated concepts.

3. Final Answer:

   To group related data logically inside another structure -> Option D

4. Quick Check:

   Nested structures = group data logically [OK]

Hint: Think of nested structures as folders inside folders [OK]

Common Mistakes:

• Confusing nested structures with inheritance
• Thinking nested structures increase structure size unnecessarily
• Assuming nested structures add functions

2. Which of the following is the correct syntax to access member city inside a nested structure address within person?

easy

A. person->address.city

B. person.address.city

C. person.address->city

D. person->address->city

Solution

1. Step 1: Identify access operator for nested structures

   When accessing members of nested structures by variable (not pointer), use dot operator for each level.

2. Step 2: Apply to given example

   Since person is a variable, access address with dot, then city with dot: person.address.city.

3. Final Answer:

   person.address.city -> Option B

4. Quick Check:

   Use dot for nested struct members when variable, not pointer [OK]

Hint: Use dot operator for nested struct members if not pointers [OK]

Common Mistakes:

• Using arrow operator (->) when variable is not a pointer
• Mixing dot and arrow operators incorrectly
• Trying to access nested member directly without parent

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

struct Date {
  int day;
  int month;
};
struct Event {
  char name[10];
  Date date;
};

int main() {
  Event e = {"Party", {25, 12}};
  std::cout << e.name << " on " << e.date.day << "/" << e.date.month << std::endl;
  return 0;
}

medium

A. Party on 25/12

B. Party on 12/25

C. Compilation error

D. Party on 0/0

Solution

1. Step 1: Understand structure initialization

   Event e is initialized with name "Party" and Date with day=25, month=12.

2. Step 2: Trace output statement

   Output prints e.name, then e.date.day (25), then e.date.month (12) separated by slashes.

3. Final Answer:

   Party on 25/12 -> Option A

4. Quick Check:

   Nested struct initialized correctly = Party on 25/12 [OK]

Hint: Match initialization order with nested structure fields [OK]

Common Mistakes:

• Swapping day and month values
• Assuming compilation error due to nested struct
• Forgetting to use dot operator to access nested members

4. Identify the error in the following code snippet:

struct Point {
  int x, y;
};
struct Rectangle {
  Point topLeft;
  Point bottomRight;
};

int main() {
  Rectangle rect;
  rect.topLeft->x = 0;
  rect.topLeft->y = 10;
  return 0;
}

medium

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

B. Missing semicolon after struct definitions

C. Incorrect member names in Rectangle

D. Rectangle struct cannot contain Point structs

Solution

1. Step 1: Check member access operators

   rect.topLeft is a structure variable, not a pointer, so dot operator (.) must be used, not arrow (->).

2. Step 2: Identify error cause

   Using -> on non-pointer causes compilation error.

3. Final Answer:

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

4. Quick Check:

   Use dot for struct variables, arrow for pointers [OK]

Hint: Use dot for struct variables, arrow only for pointers [OK]

Common Mistakes:

• Confusing dot and arrow operators
• Assuming arrow works for all struct members
• Ignoring compiler error messages

5. Given the following nested structures, how would you write a function printBookInfo that prints the book title and the author's city?

struct Author {
  char name[50];
  char city[30];
};
struct Book {
  char title[100];
  Author author;
};

hard

A. void printBookInfo(Book b) { std::cout << b.title << ", " << b.author->city << std::endl; }

B. void printBookInfo(Book b) { std::cout << b.title << ", " << b.city << std::endl; }

C. void printBookInfo(Book b) { std::cout << b.title << ", " << b.author.city << std::endl; }

D. void printBookInfo(Book b) { std::cout << b->title << ", " << b->author.city << std::endl; }

Solution

1. Step 1: Understand structure member access

   Book variable b has member title and nested structure author. Access author.city with dot operator.

2. Step 2: Check each option

   void printBookInfo(Book b) { std::cout << b.title << ", " << b.author.city << std::endl; } correctly uses b.title and b.author.city with dot operators. Others misuse arrow or omit author.

3. Final Answer:

   void printBookInfo(Book b) { std::cout << b.title << ", " << b.author.city << std::endl; } -> Option C

4. Quick Check:

   Access nested struct members with dot: b.author.city [OK]

Hint: Use dot operator for nested struct members in function parameters [OK]

Common Mistakes:

• Using arrow operator on non-pointer parameter
• Skipping nested struct name when accessing member
• Using pointer syntax incorrectly