Call stack behavior in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of nested function calls with stack frames

What is the output of the following C++ program that uses nested function calls?

void funcC() {
    std::cout << "C ";
}

void funcB() {
    std::cout << "B ";
    funcC();
}

void funcA() {
    std::cout << "A ";
    funcB();
}

int main() {
    funcA();
    return 0;
}

C++

void funcC() {
    std::cout << "C ";
}

void funcB() {
    std::cout << "B ";
    funcC();
}

void funcA() {
    std::cout << "A ";
    funcB();
}

int main() {
    funcA();
    return 0;
}

A A B C

B C B A

C A C B

D B A C

💡 Hint

Think about the order in which functions are called and how the call stack works.

The main function calls funcA, which prints 'A ' and calls funcB. funcB prints 'B ' and calls funcC, which prints 'C '. The output is 'A B C '.

❓ Predict Output

intermediate

Value of variable after recursive calls

What is the value of variable result after running this C++ recursive function?

int factorial(int n) {
    if (n <= 1) return 1;
    return n * factorial(n - 1);
}

int main() {
    int result = factorial(4);
    std::cout << result;
    return 0;
}

C++

int factorial(int n) {
    if (n <= 1) return 1;
    return n * factorial(n - 1);
}

int main() {
    int result = factorial(4);
    std::cout << result;
    return 0;
}

A 16

B 24

C 10

D 120

💡 Hint

Recall how factorial is calculated recursively.

factorial(4) = 4 * factorial(3) = 4 * 3 * factorial(2) = 4 * 3 * 2 * factorial(1) = 4 * 3 * 2 * 1 = 24.

❓ Predict Output

advanced

Call stack depth with mutual recursion

Consider the following C++ code with mutual recursion. What is the output?

void funcA(int n);
void funcB(int n);

void funcA(int n) {
    if (n == 0) return;
    std::cout << "A";
    funcB(n - 1);
}

void funcB(int n) {
    if (n == 0) return;
    std::cout << "B";
    funcA(n / 2);
}

int main() {
    funcA(4);
    return 0;
}

C++

void funcA(int n);
void funcB(int n);

void funcA(int n) {
    if (n == 0) return;
    std::cout << "A";
    funcB(n - 1);
}

void funcB(int n) {
    if (n == 0) return;
    std::cout << "B";
    funcA(n / 2);
}

int main() {
    funcA(4);
    return 0;
}

A ABABB

B ABABA

C ABAB

D ABA

💡 Hint

Trace the calls carefully, noting how n changes.

funcA(4) prints 'A' then calls funcB(3). funcB(3) prints 'B' then calls funcA(1). funcA(1) prints 'A' then calls funcB(0). funcB(0) returns. So output is 'ABA'.

❓ Predict Output

advanced

Stack unwinding with exception handling

What is the output of this C++ program demonstrating stack unwinding with exceptions?

void func3() {
    std::cout << "func3 start\n";
    throw std::runtime_error("error");
    std::cout << "func3 end\n";
}

void func2() {
    std::cout << "func2 start\n";
    func3();
    std::cout << "func2 end\n";
}

void func1() {
    std::cout << "func1 start\n";
    try {
        func2();
    } catch (...) {
        std::cout << "Exception caught in func1\n";
    }
    std::cout << "func1 end\n";
}

int main() {
    func1();
    return 0;
}

C++

void func3() {
    std::cout << "func3 start\n";
    throw std::runtime_error("error");
    std::cout << "func3 end\n";
}

void func2() {
    std::cout << "func2 start\n";
    func3();
    std::cout << "func2 end\n";
}

void func1() {
    std::cout << "func1 start\n";
    try {
        func2();
    } catch (...) {
        std::cout << "Exception caught in func1\n";
    }
    std::cout << "func1 end\n";
}

int main() {
    func1();
    return 0;
}

A

func1 start
func2 start
func3 start
Exception caught in func1
func1 end

B

func1 start
func2 start
func3 start
func3 end
func2 end
func1 end

C

func1 start
func2 start
Exception caught in func1
func1 end

D

func1 start
func2 start
func3 start
func2 end
func1 end

💡 Hint

Remember that code after throw is not executed and stack unwinds to catch block.

func3 prints 'func3 start' then throws an exception, so 'func3 end' is skipped. func2 does not print 'func2 end' because func3 threw. The exception is caught in func1, which prints the catch message and then 'func1 end'.

❓ Predict Output

expert

Call stack size after deep recursion

What is the output of this C++ program that counts recursive calls?

int count = 0;

void recurse(int n) {
    if (n == 0) return;
    count++;
    recurse(n - 1);
}

int main() {
    recurse(1000);
    std::cout << count;
    return 0;
}

C++

int count = 0;

void recurse(int n) {
    if (n == 0) return;
    count++;
    recurse(n - 1);
}

int main() {
    recurse(1000);
    std::cout << count;
    return 0;
}

A 1001

B 999

C 1000

D 0

💡 Hint

Count increments before each recursive call until n reaches zero.

The function increments count for each call except when n is zero. Since recurse(1000) calls recurse(999) ... recurse(1), count increments 1000 times.