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.

Practice

1. What does the call stack do when a function is called in C++?

easy

A. It adds a new frame for the function call on top of the stack.

B. It removes the last function call from the stack.

C. It clears all previous function calls.

D. It pauses the program execution permanently.

Solution

1. Step 1: Understand function call behavior

   When a function is called, the program needs to remember where to return after the function finishes.

2. Step 2: Role of the call stack

   The call stack adds a new frame on top to store this information and local variables.

3. Final Answer:

   It adds a new frame for the function call on top of the stack. -> Option A

4. Quick Check:

   Function call = add frame [OK]

Hint: Call means add frame; return means remove frame [OK]

Common Mistakes:

• Thinking the stack removes frames on function call
• Believing the stack clears all calls at once
• Confusing call stack with program pause

2. Which of the following C++ code snippets correctly shows a function call that affects the call stack?

easy

A. int main() { return 0; } void foo() { foo(); }

B. void foo() { main(); } int main() { return 0; }

C. void foo() {} int main() { foo(); return 0; }

D. int main() { int x = 5; }

Solution

1. Step 1: Identify valid function call from main

   void foo() {} int main() { foo(); return 0; } calls foo() from main, which adds a frame to the call stack.

2. Step 2: Check other options for correctness

   void foo() { main(); } int main() { return 0; } calls main() from foo(), which is invalid as main should not be called explicitly. int main() { return 0; } void foo() { foo(); } has foo calling itself but foo is not called from main, so no stack change from main. int main() { int x = 5; } has no function calls.

3. Final Answer:

   void foo() {} int main() { foo(); return 0; } -> Option C

4. Quick Check:

   Valid call from main = void foo() {} int main() { foo(); return 0; } [OK]

Hint: Look for main calling another function directly [OK]

Common Mistakes:

• Thinking main can be called like a normal function
• Ignoring function call order
• Confusing declaration with call

3. What is the output of this C++ program?

#include <iostream>

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

int main() {
  print(3);
  return 0;
}

medium

A. 3 2 1

B. 1 2 3

C. 0 1 2 3

D. 3 2 1 0

Solution

1. Step 1: Trace recursive calls

   print(3) calls print(2), which calls print(1), which calls print(0). At print(0), it returns immediately.

2. Step 2: Print after returning from recursion

   After print(0) returns, print(1) prints '1 ', then print(2) prints '2 ', then print(3) prints '3 '.

3. Final Answer:

   1 2 3 -> Option B

4. Quick Check:

   Recursion prints after calls = 1 2 3 [OK]

Hint: Print happens after recursion call returns [OK]

Common Mistakes:

• Printing before recursive call instead of after
• Including 0 in output incorrectly
• Confusing order of prints in recursion

4. The following code causes a runtime error. What is the main cause?

void recurse(int n) {
  if (n > 0) {
    recurse(n + 1);
  }
}

int main() {
  recurse(1);
  return 0;
}

medium

A. Stack overflow due to infinite recursion increasing n.

B. Syntax error because of missing return statement.

C. No error; program runs and exits normally.

D. Variable n is not initialized.

Solution

1. Step 1: Analyze recursion condition

   The function calls recurse(n + 1) if n > 0, so n increases each call without stopping.

2. Step 2: Understand call stack effect

   Because n never reaches 0 or less, recursion never stops, causing stack frames to pile up until overflow.

3. Final Answer:

   Stack overflow due to infinite recursion increasing n. -> Option A

4. Quick Check:

   Infinite recursion = stack overflow [OK]

Hint: Check if recursion moves toward base case [OK]

Common Mistakes:

• Assuming recursion stops without base case
• Confusing syntax error with runtime error
• Ignoring infinite growth of parameter

5. Given this code, what will be the final value of result?

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

int main() {
  int result = factorial(4);
  return 0;
}

hard

A. 1

B. 10

C. 4

D. 24

Solution

1. Step 1: Understand factorial recursion

   factorial(4) calls factorial(3), factorial(3) calls factorial(2), factorial(2) calls factorial(1), which returns 1.

2. Step 2: Calculate return values step-by-step

   factorial(2) = 2 * 1 = 2; factorial(3) = 3 * 2 = 6; factorial(4) = 4 * 6 = 24.

3. Final Answer:

   24 -> Option D

4. Quick Check:

   4! = 24 [OK]

Hint: Multiply down from n to 1 recursively [OK]

Common Mistakes:

• Stopping recursion too early
• Adding instead of multiplying
• Confusing base case return value