Accessing structure members in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of accessing structure members with dot operator

What is the output of this C++ code snippet?

C++

struct Point {
    int x;
    int y;
};

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

A Compilation error

B 3,4

C 4,3

D x=3,y=4

💡 Hint

Remember the dot operator accesses members directly by name.

The structure Point has members x and y. The code initializes p.x=3 and p.y=4, so printing p.x and p.y outputs '3,4'.

❓ Predict Output

intermediate

Output when accessing structure members via pointer

What will this C++ program print?

C++

struct Data {
    int value;
};

int main() {
    Data d = {10};
    Data* ptr = &d;
    std::cout << ptr->value << std::endl;
    return 0;
}

A Garbage value

B Compilation error

C 10

D 0

💡 Hint

Use the arrow operator to access members through pointers.

The pointer ptr points to d, which has value 10. Using ptr->value accesses that member correctly.

❓ Predict Output

advanced

Output of nested structure member access

What is the output of this code?

C++

struct Inner {
    int a;
};

struct Outer {
    Inner inner;
};

int main() {
    Outer o = {{5}};
    std::cout << o.inner.a << std::endl;
    return 0;
}

A 5

B Compilation error

C 0

D Garbage value

💡 Hint

Access nested members by chaining dot operators.

The Outer structure contains Inner named inner. The value a is initialized to 5 and printed.

❓ Predict Output

advanced

Output when accessing structure members with pointer and dot

What happens when you run this code?

C++

struct Item {
    int id;
};

int main() {
    Item item = {7};
    Item* p = &item;
    std::cout << (*p).id << std::endl;
    return 0;
}

A 7

B Compilation error

C 0

D Runtime error

💡 Hint

Dereference pointer first, then use dot operator.

(*p).id accesses the id member of the object pointed to by p, which is 7.

🧠 Conceptual

expert

Error type when accessing structure member incorrectly

What error will this code produce?

C++

struct Sample {
    int x;
};

int main() {
    Sample* s = nullptr;
    std::cout << s->x << std::endl;
    return 0;
}

A Compilation error

B Undefined behavior but no crash

C Output: 0

D Segmentation fault (runtime error)

💡 Hint

Accessing member through null pointer causes crash.

Dereferencing a null pointer to access a member causes a segmentation fault at runtime.

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