Passing parameters by reference in C++

Introduction

Passing parameters by reference lets a function change the original value of a variable. This saves memory and can make programs faster.

When you want a function to update the original variable, like changing a score in a game.

When you want to avoid copying large data, like big lists or objects, to save memory.

When you want multiple outputs from a function by changing several variables.

When you want to improve performance by not copying data unnecessarily.

Syntax

C++

void functionName(type &parameterName) {
    // code that uses parameterName
}

The ampersand (&) after the type means the parameter is passed by reference.

Changes to the parameter inside the function affect the original variable.

Examples

This function adds 1 to the original integer passed in.

C++

void increase(int &num) {
    num = num + 1;
}

This function swaps the values of two integers by reference.

C++

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

Sample Program

This program shows how the function addFive changes the original variable value by adding 5 to it.

C++

#include <iostream>
using namespace std;

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

int main() {
    int value = 10;
    cout << "Before: " << value << endl;
    addFive(value);
    cout << "After: " << value << endl;
    return 0;
}

Important Notes

Passing by reference allows the function to modify the original variable directly.

Be careful: changes inside the function affect the original data.

If you don't want the function to change the variable, use const reference.

Summary

Passing parameters by reference lets functions change original variables.

Use & after the type in the function parameter to pass by reference.

This method saves memory and can improve performance.

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