Stack recursion Practice question

Duration: 2 min

This video lesson is available to enrolled students.

Enroll to watch — ISRO Scientist/Engineer 'SC'

AI Summary

An AI-generated summary of this video lecture.

The video presents a C programming problem involving a recursive function named fun. The code snippet includes standard headers and a function definition where fun(int n) checks if n equals 4. If true, it returns n; otherwise, it returns 2 * fun(n+1). The main function calls fun(2) and prints the result. The instructor systematically traces the execution flow on a whiteboard to determine the final output value, demonstrating the mechanics of recursion and stack unwinding.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The instructor begins the analysis by writing fun(2) on the whiteboard, representing the initial function call from main. He points to the if (n == 4) condition in the code, explaining that since n is 2, the condition is false. This directs execution to the else block, which returns 2 * fun(n+1). He writes return 2 * fun(3) on the board. Next, he traces the call fun(3). Again, n is not 4, so he writes return 2 * fun(4). For the call fun(4), the condition n == 4 is true, so the function returns the value 4. He writes fun(4) = 4 on the side. He then substitutes this value back up the stack: fun(3) becomes 2 * 4 = 8. Finally, fun(2) becomes 2 * 8 = 16. He writes (16) and marks option (C) as correct, crossing out the other choices.

  2. 2:00 – 2:18 02:00-02:18

    The video concludes with the final state of the whiteboard visible. The recursive expansion is complete, showing the hierarchy of calls: fun(2) depends on fun(3), which depends on fun(4). The calculated intermediate values 4, 8, and the final result 16 are clearly written in red ink. The instructor has placed a checkmark next to option (C) 16, confirming the solution. The screen displays the multiple-choice options with the correct answer highlighted, ending the lecture segment.

This lecture effectively teaches the method of tracing recursive code by expanding the call stack and substituting return values. It emphasizes identifying the base case to stop recursion and correctly computing the final result through sequential multiplication.