Question

Write a function that takes an int as an argument and returns true if the int is an even number and false if not. In any case change the value of the original argument to twice its value. Use the integers 0 through 4 as test cases. Do not use any arrays.

Answer 1

Code:

#include <iostream>
using namespace std;
bool Check(int number) //Check() fucntion which takes a number as an argumnet
{
  
if (number%2==0) //if a number is divisible by 2 i. even then return true and print the twice of the given nuumber
{
cout << "Twice of the original value " << number << " is " << 2*number << endl;
return true;
}
else{ //else print the twice of the given number and return false
cout << "Twice of the original value " << number << " is " << 2*number << endl;
return false;
}
}
int main()
{

int n;
cout << "Enter a number: "; //asking user to enter a number
cin >> n;
int val=Check(n); //calling Check() function by passing n value and storing the result in val
if (val==0)
cout << "False"; //if val is 0 print false
else
cout << "True"; //else print True
return 0;
}
Code and Output Screenshots:

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