C program for odd or even?

Answer 1

# include

void main()

{

int a;

printf("\n Enter a number to find if its odd or even");

scanf("%d",&a);

if(a%2==0)

{

printf("\n The number entered is even");

}

else

{

printf("\n The number entered is odd");

}

getch();

}

Answer 2

Write a program that inputs a series of integers and passes them one at a time to a function even which uses the remainder operator to determine if an integer is even. The function should take an integer argument and return 1 if the integer is even and 0 otherwise.

Answer 3

Add one to (a copy of) the number and then take the integer of the result. If the two values are the same, then the original number was even. (This only works correctly for integers.)

Answer 4

I will not write a program for you submit as your homework.

A integer number can checked for odd or even by looking at the low order bit. If the low order bit is 1 (true) the number is odd.

The rest is up to you.

Answer 5

#include

using std::cin;
using std::cout;

int main()
{
int num = 0;
cout << endl << "Enter a number: ";
cin >> num;

if (num % 2 == 0)
{
cout << endl << "The number " << num << " is even";

}
else
{
cout << endl << "The number " << num << " is odd";

}

system("PAUSE");
return 0;

}

Answer 6

Considering you already got the number input from keyboard or file, and "int a" will be your number.

if (a/2 == 0)

{

cout << a << "is even"

}

else

{

cout << a << "is odd"

}

Answer 7

When we divide any whole number (integer) by 2, the remainder can only be 0 or 1. If the remainder is 0 then the number must be even, otherwise it must be odd. Given that 0 and 1 implicitly convert to the Boolean values false and true, respectively, we can easily implement the required functions as follows, using the modulo operator (%) to determine the remainder after integer division by 2:

bool is_odd (int num) { return num%2; } // means: (num%2)==1
bool is_even (int num) { return !is_odd (num); } // means: (num%2)!=1 or (num%2)==0;

Note that for unsigned integers, we can improve efficiency further by testing the low-order (least-significant) bit using the bitwise logical AND operator (&):

bool is_odd (unsigned num) { return num&1; } // means: num&1==1;
bool is_even (unsigned num) { return !is_odd(num); } // means: (num&1)!=1 or (num&1)==0;

This works because the least-significant bit of an unsigned integer represents the value 2^0 which is 1 (hence we test for 1), while all the higher-order bits represent increasing powers of 2 which are 2, 4, 8, etc, all of which are even. So, regardless of which other bits are set (or not), the low-order bit alone tells us if the number is odd or even. If it is set, the number is odd, otherwise it is even.

Although bitwise logic is much faster than modulo integer division, it is not guaranteed to be portable for all signed integers given that some (older) systems represent negative integers using ones-complement notation while most (newer) systems use twos-complement notation, thus complicating the logic. The first solution shown above is guaranteed to be portable across all systems whereas testing for and converting negative values to positive values to take advantage of the bitwise logic in the second solution would effectively negate that advantage. Therefore use the second solution only when all integers to be tested are guaranteed to be unsigned and therefore do not require conversion, or if the systems are guaranteed to be twos-complement systems.

Answer 8

int var;
...
if (var & 1) printf("%d is odd\n", var);
else printf("%d is even\n", var);

Answer 9

int isEven (int x) return x % 2 == 0 ? 1 : 0;