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elena55 [62]
3 years ago
5

Write a program segment with a do-while loop that displays whether a user-entered integer is even or odd. The code should then a

sk the user if he or she wants to test another number. The loop should repeat as long as the user enters 'Y' or 'y' . Use a logical OR operator in the do-while loop test expression
Computers and Technology
1 answer:
Leya [2.2K]3 years ago
5 0

Answer:

Following is the program in C++ program

#include<iostream> // header file

using namespace std; // namespace  

int main() // main function  

{

int num1 = 0; // variable declaration  

char test;// variable declaration  

do

{

cout << " Enter the number: ";

cin >> num1; //Read the input by the user  

if (num1 % 2 == 0) // check the condition of even

cout << " number is even.\n";

else // check the condition of odd

cout << "Number is odd.\n";

cout << " Do you wanted the another test of number (y/n)? ";

cin >> test; //Read the input by user

} while (test == 'y' || test == 'Y'); //terating the loop

return 0;    

}

Output:

Enter the number:45

number is even.

Do you wanted the another test of number (y/n) Y

Enter the number:5

Number is odd.

Do you wanted the another test of number (y/n) n

Explanation:

Following are the description of program

  • Declared a variable "num1" as integer type and initialized as 0 to them
  • Declared a variable "test" as char type .
  • After that iterating the do -while loop .Read the value by the user in the "num1" variable .
  • Check the condition of even and odd by using % operator .
  • Read the value of test by the user .If the user enter Y or y then loop again executing otherwise not .
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