CPP Exception Handling Hackerrank Solution in C++

Cpp Exception Handling Hackerrank Solution in C++. In this challenge, the task is to debug the existing code to successfully execute all provided test files. You are required to extend the existing code so that it handles the std::invalid_argument exception properly. More specifically, you have to extend the implementation of the process_input function. It takes integer n as an argument and has to work as follows:

It calls function largest_proper_divisor(n).
If this call returns a value without raising an exception, it should print in a single line result=d where d is the returned value.
Otherwise, if the call raises a std::invalid_argument exception, it has to print in a single line the string representation of the raised exception, i.e. its message.
Finally, no matter if the exception is raised or not, it should print in a single line returning control flow to the caller after any other previously printed output.

To keep the code quality high, you are advised to have exactly one line printing returning control flow to the caller in the body of the process_input function.

Your function will be tested against several cases by the locked template code.

Cpp Exception Handling Hackerrank Solution in C++

Input Format

The input is read by the provided locked code template. In the only line of the input, there is a single integer n, which is going to be the argument passed to function process_input.

Constraints

0 <= n <= 100

Output Format

The output should be produced by the function process_input as described in the statement.

Sample Input 0

Sample Output 0

the largest proper divisor is not defined for n=0

returning control flow to the caller

Explanation 0

In the first sample, n = 0, so the call largest_proper_divisor(0) raises an exception. In this case, the function process_input prints two lines. In the first of them, it prints the string representation of the raised exception, and in the second line, it prints returning control flow to the caller.

Sample Input 1

Sample Output 1

result=3

returning control flow to the caller

Explanation 1

In the first sample, n = 9, so the call largest_proper_divisor(9) doesn't raise an exception and returns value 3. In this case, the function process_input prints two lines. In the first of them, it prints result=3 because the returned value by largest_proper_divisor(9) is 3, and in the second line it prints returning control flow to the caller.

Submit your solution here: Click here

Cpp Exception Handling Hackerrank Solution in C++

#include <iostream>
#include <stdexcept>
using namespace std;

int largest_proper_divisor(int n)
{
	if (n == 0)
	{
		throw invalid_argument("largest proper divisor is not defined for n=0");
	}

	if (n == 1)
	{
		throw invalid_argument("largest proper divisor is not defined for n=1");
	}

	for (int i = n / 2; i >= 1; --i)
	{
		if (n % i == 0)
		{
			return i;
		}
	}

	return -1;	// will never happen
}

void process_input(int n)
{
	int d;
	try
	{
		d = largest_proper_divisor(n);
		cout << "result=" << d << endl;
	}

	catch (invalid_argument & ex)
	{
		cout << ex.what() << endl;
	}

	cout << "returning control flow to caller";
}

int main()
{
	int n;
	cin >> n;
	process_input(n);
	return 0;
}