Chapter 17: Problem 13
Consider the following function: int func(int x) { if (x == 0) return 2; else if (x == 1) return 3; else return (func(x - 1) + func(x - 2)); } What is the output of the following statements? a. cout << func(0) << endl; b. cout << func(1) << endl; c. cout << func(2) << endl; d. cout << func(5) << endl;
Short Answer
Step by step solution
Understanding the Function
Calculate func(0)
Calculate func(1)
Calculate func(2)
Calculate func(5)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Recursion
To understand recursion, it's crucial to visualize the call stack, where each function call is stacked until a condition stops further calls. This stopping condition is known as the base case. Recursion is particularly useful for problems that exhibit optimal substructure and can be defined in terms of themselves, such as the classic Fibonacci sequence.
Function Calls
In the context of the recursive `func` function, every time it calls itself, a new function call is added to the call stack. This process continues until the function encounters a base case—either `x == 0` or `x == 1`. Each call to a function returns a value that will be used in calculations higher in the stack, until the original function call finally returns the total value to the main program. Remember to ensure that all function calls eventually lead to a base case to prevent infinite recursion.
Base Case
In the `func` function from the exercise, there are two base cases:
- If `x == 0`, the function returns 2.
- If `x == 1`, the function returns 3.
C++ Programming
When writing recursive functions in C++, keep these tips in mind:
- Always define a base case to prevent infinite recursion and stack overflow errors.
- Consider using memoization to improve performance if the same results are computed multiple times.
- Be mindful of memory and performance issues, as recursion can be costly in terms of resources for deep recursive calls.