Accessing structure members in C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a structure in C++?

A structure in C++ is a user-defined data type that groups related variables of different types under one name.

beginner

How do you access a member of a structure variable in C++?

You use the dot operator (.) to access a member of a structure variable. For example, if person is a structure variable and age is a member, you write person.age.

beginner

How do you access a member of a structure through a pointer in C++?

You use the arrow operator (->) to access a member of a structure through a pointer. For example, if pPerson is a pointer to a structure and age is a member, you write pPerson->age.

intermediate

What is the difference between the dot operator (.) and the arrow operator (->) when accessing structure members?

The dot operator (.) is used to access members of a structure variable directly, while the arrow operator (->) is used to access members through a pointer to a structure.

beginner

Given the structure struct Point { int x; int y; }; and a variable Point p = {3, 4};, how do you print the value of x?

You print the value of x using p.x. For example: std::cout << p.x;

Which operator is used to access a member of a structure variable directly?

A . (dot)

B -> (arrow)

C * (asterisk)

D & (ampersand)

The dot operator (.) is used to access members of a structure variable directly.

How do you access a member of a structure through a pointer?

A . (dot)

B -> (arrow)

C :: (scope resolution)

D # (preprocessor)

The arrow operator (->) is used to access members through a pointer to a structure.

Given struct Car { int speed; }; and Car* pCar;, how do you access the speed member?

A &pCar.speed

B pCar.speed

C *pCar.speed

D pCar->speed

You use the arrow operator with pointers: pCar->speed.

If person is a structure variable, which is correct to access its member name?

A person->name

B *person.name

C person.name

D &person->name

Use the dot operator to access members of a structure variable: person.name.

What will happen if you try to use the dot operator on a pointer to a structure?

A It will cause a compilation error.

B It will access the member correctly.

C It will access the member but with wrong value.

D It will convert the pointer to a variable automatically.

Using the dot operator on a pointer to a structure causes a compilation error; you must use the arrow operator.

Explain how to access members of a structure variable and a pointer to a structure in C++.

Think about the difference between a variable and a pointer.

Describe the difference between the dot operator and the arrow operator when working with structures.

Consider how pointers and variables behave differently.

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