Cpp exception handling in C++ HackerRank Solution

Hello Programmers, In this post, you will know how to solve the Cpp exception handling in C++ HackerRank Solution. This problem is a part of the HackerRank C++ Programming Series.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

Cpp exception handling in C++ HackerRank Solution

Problem

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 std::invalid_argument exception properly. More specifically, you have to extend the implementation of process_input function. It takes integer n as an argument and has to work as follows:

1. It calls function largest_proper_divisor(n).
2. 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.
3. 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.
4. Finally, no matter if the exception is raised or not, it should print in a single line returning control flow to 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 caller in the body of process_input function.

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

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 function process_input as described in the statement.

Sample Input 0:

0

Sample Output 0:

largest proper divisor is not defined for n=0
returning control flow to 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 caller.

Sample Input 1:

9

Sample Output 1:

result=3
returning control flow to 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 caller.

Cpp exception handling in C++ HackerRank Solution

#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
}
/* Cpp exception handling in C++ - Hacker Rank Solution START */
void process_input(int n)
{
    try{
    int d = largest_proper_divisor(n);
    cout << "result=" << d << endl;
    }
    catch(invalid_argument& e)
    {
        cout << e.what() << endl;
    }
    cout << "returning control flow to caller" << endl;
}
/*Cpp exception handling in C++ - Hacker Rank Solution END */
int main()
{
    int n;
    cin >> n;
    process_input(n);
    return 0;
}

Disclaimer: The above Problem (Cpp exception handling in C++) is generated by Hackerrank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

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