Concept explainers
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)”...
- Here, the value of “n” is not less than or not equal to “1”. So, performs “else” statement.
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