Passing parameters by reference in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of modifying a variable passed by reference

What is the output of this C++ program?

C++

#include <iostream>

void addFive(int &num) {
    num += 5;
}

int main() {
    int x = 10;
    addFive(x);
    std::cout << x << std::endl;
    return 0;
}

A 10

B 15

C 5

D Compilation error

💡 Hint

Remember that passing by reference allows the function to modify the original variable.

The function addFive takes an int reference and adds 5 to it. Since x is passed by reference, its value changes from 10 to 15.

❓ Predict Output

intermediate

Effect of passing pointer vs reference

What will be printed by this program?

C++

#include <iostream>

void increment(int *ptr) {
    (*ptr)++;
}

void incrementRef(int &ref) {
    ref++;
}

int main() {
    int a = 3;
    increment(&a);
    incrementRef(a);
    std::cout << a << std::endl;
    return 0;
}

A 3

B 4

C Compilation error

D 5

💡 Hint

Both functions increase the value by one.

First, increment increases a from 3 to 4 using a pointer. Then incrementRef increases it from 4 to 5 using a reference.

🔧 Debug

advanced

Identify the error with passing by reference

What error will this code produce?

C++

#include <iostream>

void setToZero(int &num) {
    num = 0;
}

int main() {
    setToZero(5);
    return 0;
}

A Compilation error: cannot bind non-const lvalue reference to an rvalue

B Runtime error: segmentation fault

C No error, program runs fine

D Output: 0

💡 Hint

You cannot pass a literal to a non-const reference parameter.

The function expects a reference to a variable, but 5 is a literal (rvalue). This causes a compilation error.

❓ Predict Output

advanced

Reference parameter with const qualifier

What is the output of this program?

C++

#include <iostream>

void printValue(const int &num) {
    // num += 1; // Uncommenting this line causes error
    std::cout << num << std::endl;
}

int main() {
    int x = 7;
    printValue(x);
    return 0;
}

A 7

B 8

C Compilation error due to modification of const reference

D Runtime error

💡 Hint

Const reference means the function cannot change the value.

The function prints the value of x, which is 7. The commented line would cause a compilation error if uncommented.

❓ Predict Output

expert

Reference to pointer parameter behavior

What will be the output of this program?

C++

#include <iostream>

void changePointer(int *&ptr) {
    static int y = 20;
    ptr = &y;
}

int main() {
    int x = 10;
    int *p = &x;
    changePointer(p);
    std::cout << *p << std::endl;
    return 0;
}

A Compilation error

B 10

C 20

D Undefined behavior

💡 Hint

The function changes the pointer itself to point to a different variable.

The function takes a reference to a pointer and changes it to point to y (20). So *p prints 20.

Practice

1.

What does passing a parameter by reference in C++ allow a function to do?

easy

A. Only read the variable without access to its value

B. Create a copy of the variable inside the function

C. Prevent the variable from being changed

D. Modify the original variable passed to the function

Solution

1. Step 1: Understand passing by reference

   Passing by reference means the function receives the actual variable's address, not a copy.

2. Step 2: Effect on the original variable

   Since the function works with the original variable, any changes inside the function affect it directly.

3. Final Answer:

   Modify the original variable passed to the function -> Option D

4. Quick Check:

   Passing by reference = modify original [OK]

Hint: Reference means function can change original variable [OK]

Common Mistakes:

• Thinking it creates a copy instead of using original
• Confusing with passing by value
• Assuming it prevents changes

2.

Which of the following is the correct syntax to pass an integer num by reference to a function update?

void update(??? num) { num = 10; }

easy

A. int &num

B. int *num

C. int num

D. int &&num

Solution

1. Step 1: Identify reference syntax in C++

   To pass by reference, use & after the type in the parameter list.

2. Step 2: Match syntax to options

   int &num uses int &num, which is the correct way to pass by reference.

3. Final Answer:

   int &num -> Option A

4. Quick Check:

   Reference parameter uses & after type [OK]

Hint: Use & after type to pass by reference [OK]

Common Mistakes:

• Using pointer syntax (*) instead of reference (&)
• Omitting & and passing by value
• Confusing rvalue reference (&&) with normal reference

3.

What is the output of the following code?

void addOne(int &x) {
    x = x + 1;
}

int main() {
    int a = 5;
    addOne(a);
    std::cout << a << std::endl;
    return 0;
}

medium

A. Compilation error

B. 5

C. 6

D. Undefined behavior

Solution

1. Step 1: Understand function behavior

   The function addOne takes x by reference and increments it by 1.

2. Step 2: Trace the variable a

   Initially, a is 5. After calling addOne(a), a becomes 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Reference parameter increments original variable [OK]

Hint: Reference changes original variable value [OK]

Common Mistakes:

• Thinking original variable stays unchanged
• Confusing reference with pointer syntax
• Expecting compilation error due to missing &

4.

Identify the error in this code snippet:

void swap(int &a, int &b) {
    int temp = a;
    a = b;
    b = temp;
}

int main() {
    int x = 10, y = 20;
    swap(x, y);
    std::cout << x << ", " << y << std::endl;
    return 0;
}

medium

A. Missing & in function parameters

B. No error, code works correctly

C. Incorrect swapping logic

D. Missing return type in main

Solution

1. Step 1: Check function parameter syntax

   The function swap correctly uses references int &a and int &b.

2. Step 2: Verify swapping logic and main function

   The swapping logic uses a temporary variable correctly. The main function has proper return type and prints swapped values.

3. Final Answer:

   No error, code works correctly -> Option B

4. Quick Check:

   Correct reference swap function = no error [OK]

Hint: Check reference syntax and logic carefully [OK]

Common Mistakes:

• Assuming swap logic is wrong
• Confusing pointers with references
• Thinking main lacks return type

5.

Given this function that doubles the value of an integer passed by reference, what will be the output of the program?

void doubleValue(int &num) {
    num *= 2;
}

int main() {
    int values[] = {1, 2, 3};
    for (int i = 0; i < 3; i++) {
        doubleValue(values[i]);
    }
    for (int v : values) {
        std::cout << v << " ";
    }
    return 0;
}

hard

A. 2 4 6

B. 1 2 3

C. Compilation error due to reference in array

D. 0 0 0

Solution

1. Step 1: Understand passing array elements by reference

   Each element values[i] is passed by reference to doubleValue, so the original array elements are modified.

2. Step 2: Trace the loop and doubling effect

   Loop runs 3 times, doubling each element: 1->2, 2->4, 3->6.

3. Final Answer:

   2 4 6 -> Option A

4. Quick Check:

   Reference parameter doubles original array elements [OK]

Hint: Array elements passed by reference get changed directly [OK]

Common Mistakes:

• Thinking array elements are copied, not changed
• Expecting compilation error with references and arrays
• Ignoring loop effect on all elements