Passing parameters by reference in C++ - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a function parameter that passes an integer by reference.

C++

void increment([1] int &num) {
    num++;
}

Drag options to blanks, or click blank then click option'

Avolatile

Bconst

Csigned

Dunsigned

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting the & symbol to pass by reference

Using pointer syntax instead of reference

The const keyword before the reference parameter means the function cannot modify the argument. Here, we want to pass by reference, so int &num is correct, and adding const is common to prevent changes.

2fill in blank

medium

Complete the code to call the function swap passing two integers by reference.

C++

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

int main() {
    int x = 5, y = 10;
    swap([1], y);
    return 0;
}

Drag options to blanks, or click blank then click option'

Ax

B&x

C*x

Dint x

Attempts:

3 left

💡 Hint

Common Mistakes

Using &x in the function call, which is incorrect for references

Using *x which is for pointers

When passing by reference, you just pass the variable name without & or * in the call. The function parameter already expects a reference.

3fill in blank

hard

Fix the error in the function declaration to correctly pass a string by reference.

C++

#include <string>

void printName([1] std::string name) {
    std::cout << name << std::endl;
}

Drag options to blanks, or click blank then click option'

Aconst std::string &

Bstd::string *

Cstd::string &

Dstd::string

Attempts:

3 left

💡 Hint

Common Mistakes

Passing by value causes unnecessary copying

Using pointer syntax when reference is preferred

Passing a string by const std::string & is efficient and prevents modification. It avoids copying the string.

4fill in blank

hard

Fill both blanks to create a function that doubles an integer passed by reference and call it correctly.

C++

void doubleValue([1] int &num) {
    num *= 2;
}

int main() {
    int val = 7;
    doubleValue([2]);
    return 0;
}

Drag options to blanks, or click blank then click option'

Aint

Bval

Cconst

D&val

Attempts:

3 left

💡 Hint

Common Mistakes

Using &val in the call which is incorrect for references

Adding const incorrectly in the parameter

The function parameter should be int &num to pass by reference. When calling, just pass the variable val without &.

5fill in blank

hard

Fill all three blanks to create a function that swaps two integers by reference and call it correctly.

C++

void swap([1] int &a, [2] int &b) {
    int temp = a;
    a = b;
    b = temp;
}

int main() {
    int x = 3, y = 4;
    swap([3], y);
    return 0;
}

Drag options to blanks, or click blank then click option'

Aint

Bconst

Cx

D&x

Attempts:

3 left

💡 Hint

Common Mistakes

Using const in parameters which prevents modification

Passing &x in the function call which is incorrect

The function parameters should be int &a and int &b to pass by reference. When calling, just pass the variable x without &.

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