The provided code calculates the **factorial** of a given integer...

February 4, 2025 at 09:34 AM

long factorial(int n) { if (n <= 1) { // base case return 1; } else { return n * factorial(n - 1); } }

The provided code calculates the factorial of a given integer n recursively. Here's an explanation of what the code does:

  1. Base Case (Stopping Condition):

    • If n <= 1, the function returns 1. This serves as the base case to prevent infinite recursion, signifying that the factorial of 0 or 1 is 1.
  2. Recursive Call:

    • If n > 1, the function calculates n * factorial(n - 1). This means it multiplies n by the factorial of n - 1. This process will repeat until n reaches 1 (or 0), at which point the recursion stops and the intermediate results are multiplied back up the recursion chain.

Example Execution:

If n = 4, the function evaluates as follows:

  1. factorial(4) = 4 * factorial(3)
  2. factorial(3) = 3 * factorial(2)
  3. factorial(2) = 2 * factorial(1)
  4. factorial(1) returns 1 (base case).

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

Output:

This function computes the factorial of n, which is mathematically defined as: [ n! = n \times (n-1) \times (n-2) \times \dots \times 1 ]

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