Function parameters in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of function with default parameters

What is the output of this C++ code?

C++

#include <iostream>
#include <string>
using namespace std;

void greet(string name = "Guest", int times = 1) {
    for (int i = 0; i < times; i++) {
        cout << "Hello, " << name << "!\n";
    }
}

int main() {
    greet();
    greet("Alice", 2);
    return 0;
}

A

Hello, Guest!
Hello, Guest!
Hello, Alice!
Hello, Alice!

B

Hello, Guest!
Hello, Alice!
Hello, Alice!

C

Hello, Guest!
Hello, Alice!

D

Hello, Alice!
Hello, Alice!

💡 Hint

Remember how default parameters work when no arguments are passed.

The function greet has default parameters: name = "Guest" and times = 1. When called without arguments, it prints "Hello, Guest!" once. When called with "Alice" and 2, it prints "Hello, Alice!" twice.

❓ Predict Output

intermediate

Output of function with pass-by-reference parameter

What is the output of this C++ code?

C++

#include <iostream>
using namespace std;

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

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

A

5

B

Compilation error

C

6

D

Undefined behavior

💡 Hint

Think about how passing by reference affects the original variable.

The function addOne takes an integer reference and increments it. The original variable a is changed from 5 to 6, so the output is 6.

❓ Predict Output

advanced

Output of function with const reference parameter

What is the output of this C++ code?

C++

#include <iostream>
using namespace std;

void printValue(const int &val) {
    cout << val << endl;
}

int main() {
    int x = 10;
    printValue(x);
    return 0;
}

A

10

B

Compilation error

C

0

D

Undefined behavior

💡 Hint

Const references allow reading but not modifying the argument.

The function printValue takes a const reference to an int and prints it. The value of x is 10, so it prints 10.

❓ Predict Output

advanced

Output of function with pointer parameter

What is the output of this C++ code?

C++

#include <iostream>
using namespace std;

void setZero(int *ptr) {
    if (ptr) {
        *ptr = 0;
    }
}

int main() {
    int a = 5;
    setZero(&a);
    cout << a << endl;
    return 0;
}

A

0

B

5

C

Compilation error

D

Segmentation fault

💡 Hint

Pointers can be used to modify the value they point to.

The function setZero takes a pointer to int and sets the pointed value to 0. Since a is passed by address, its value changes to 0.

❓ Predict Output

expert

Output of overloaded functions with different parameter types

What is the output of this C++ code?

C++

#include <iostream>
using namespace std;

void print(int x) {
    cout << "int: " << x << endl;
}

void print(double x) {
    cout << "double: " << x << endl;
}

int main() {
    print(5);
    print(5.0);
    print('5');
    return 0;
}

A

int: 5
double: 5.0
int: 53

B

int: 5
double: 5
int: 5

C

int: 5
double: 5
Compilation error

D

int: 5
double: 5
int: 53

💡 Hint

Remember how char is promoted to int in function overloading.

The call print(5) matches print(int). The call print(5.0) matches print(double). The call print('5') passes a char, which is promoted to int with ASCII value 53, so print(int) is called printing 53.