Reference declaration in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of reference modification

What is the output of this C++ code snippet?

C++

#include <iostream>

int main() {
    int x = 10;
    int &ref = x;
    ref = 20;
    std::cout << x << std::endl;
    return 0;
}

A Compilation error

B 20

C 10

D Undefined behavior

💡 Hint

Remember that a reference is an alias to the original variable.

The reference 'ref' points to 'x'. Changing 'ref' changes 'x'. So output is 20.

❓ Predict Output

intermediate

Reference to const behavior

What will this program output?

C++

#include <iostream>

int main() {
    int x = 5;
    const int &ref = x;
    // ref = 10; // Uncommenting this line causes error
    std::cout << ref << std::endl;
    x = 15;
    std::cout << ref << std::endl;
    return 0;
}

A 5\n5

B Compilation error

C Undefined behavior

D 5\n15

💡 Hint

A const reference cannot be used to modify the variable, but the variable itself can change.

The const reference 'ref' refers to 'x'. When 'x' changes, 'ref' sees the new value.

❓ Predict Output

advanced

Reference and pointer difference

What is the output of this code?

C++

#include <iostream>

int main() {
    int a = 7;
    int &ref = a;
    int *ptr = &a;
    ref = 10;
    *ptr = 20;
    std::cout << a << std::endl;
    return 0;
}

A 7

B Compilation error

C 20

D 10

💡 Hint

Both reference and pointer modify the same variable 'a'.

First ref sets 'a' to 10, then pointer sets 'a' to 20. Final value is 20.

❓ Predict Output

advanced

Reference initialization with temporary

What error or output does this code produce?

C++

#include <iostream>

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

A Compilation error

B 5

C Runtime error

D Undefined behavior

💡 Hint

Non-const references cannot bind to temporary values.

Non-const references must bind to lvalues. '5' is an rvalue, so this is a compile error.

🧠 Conceptual

expert

Reference lifetime and dangling references

Consider this code snippet. What is the problem with it?

C++

#include <iostream>

int& getRef() {
    int x = 10;
    return x;
}

int main() {
    int &ref = getRef();
    std::cout << ref << std::endl;
    return 0;
}

A Undefined behavior due to dangling reference.

B It prints 10 correctly.

C Compilation error due to returning reference to local variable.

D Runtime error due to null reference.

💡 Hint

Local variables die when the function ends, but the reference is returned.

The function returns a reference to a local variable that no longer exists after the function ends, causing undefined behavior.