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.