Chapter 14.2, Problem 13STE

Explanation of Solution

Given code:

//Header file

#include <iostream>

//For standard input and output

using namespace std;

//Function declaration for "rose" function

int rose(int n);

//Precondition: n >= 0.

//Main function

int main()

{

  /* Call function "rose()" with argument "4" and display it */

      cout << rose(4);

      return 0;

}

/* Function definition for rose() function with parameter "n" */

int rose(int n)

{

      /* If "n" is less than or equal to "1", then */

      if (n <= 0)

            //Returns "1"

            return 1;

      //Otherwise

      else

     /* Recursively call the function "rose()" with decrement the value of "n" by "1" and multiply by "n" */

            return (rose(n - 1) * n);

}

Explanation:

The given code is used to display the value after calling the function “rose()”.

• First, declare the function for “rose()”.
• Define main function
  • Call “rose()” function with argument “4” and display the given value.
• Define “rose()” function with parameter “n”.
  • If “n” is less than or equal to “1”, then returns “1”.
  • Otherwise, recursively call the function “rose()” with decrement the value of “n” by “1” and then multiply with “n”.

Reasons for displaying given output:

• In main function, call “rose()” function with argument “4”.
• So, in the function “rose()”, first check the value of “n”.
  • Here, the value of “n” is not less than or not equal to “1”. So, performs “else” statement.
    • In “else” statement, return “(rose(4 - 1) * 4)” implies “(rose(3) * 4)”...