Call stack behavior in C++ - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to call the function foo.

C++

#include <iostream>

void foo() {
    std::cout << "Hello from foo!" << std::endl;
}

int main() {
    [1];
    return 0;
}

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

Afoo()

Bfoo

Ccall foo()

Dfoo();

Attempts:

3 left

💡 Hint

Common Mistakes

Writing only the function name without parentheses.

Adding extra words like 'call' before the function name.

To call a function in C++, you write its name followed by parentheses. So foo() calls the function.

2fill in blank

medium

Complete the code to make bar call foo.

C++

#include <iostream>

void foo() {
    std::cout << "Inside foo" << std::endl;
}

void bar() {
    [1];
}

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

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

Afoo();

Bfoo

Ccall foo()

Dfoo()

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting parentheses when calling the function.

Adding extra words like 'call' before the function name.

Inside a function, to call another function, use its name with parentheses like foo().

3fill in blank

hard

Fix the error in the recursive function countdown to avoid infinite recursion.

C++

#include <iostream>

void countdown(int n) {
    if (n == 0) {
        std::cout << "Done!" << std::endl;
        return;
    }
    std::cout << n << std::endl;
    countdown([1]);
}

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

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

A0

Bn + 1

Cn - 1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Calling with the same number causing infinite recursion.

Increasing the number instead of decreasing.

To count down, you need to call countdown with n - 1 so the number decreases each time.

4fill in blank

hard

Fill both blanks to create a dictionary-like map of numbers to their squares using a loop.

C++

#include <iostream>
#include <map>

int main() {
    std::map<int, int> squares;
    for (int [1] = 1; [2] <= 5; ++i) {
        squares[i] = i * i;
    }
    for (const auto& [key, value] : squares) {
        std::cout << key << ": " << value << std::endl;
    }
    return 0;
}

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

Ai

Bj

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in initialization and condition.

Using variables not declared in the loop.

The loop variable should be the same in initialization and condition. Here, i is used.

5fill in blank

hard

Fill all three blanks to create a recursive function that sums numbers from n down to 1.

C++

#include <iostream>

int sum_to_n(int [1]) {
    if ([2] == 0) {
        return 0;
    }
    return [3] + sum_to_n([2] - 1);
}

int main() {
    std::cout << sum_to_n(5) << std::endl;
    return 0;
}

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

An

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names causing errors.

Forgetting to decrease the number in the recursive call.

The function parameter and recursive calls use the same variable n to sum down to zero.

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