Using C++ Relational Ops to Check if a Number is Even

[ineedhelpnow]

Write an expression that will print "Even" if the value of userNum is an even number.

#include <iostream>
using namespace std;

int main() {
   int userNum = 0;

   userNum = 6;

   if (<STUDENT CODE>) { 
      cout << "Even" << endl; 
   }
   else {
      cout << "Odd" << endl;
   }

   return 0;
}

the <STUDENT CODE> is

userNum %2==0

i understand the %2 part but why equal to 0? how does the ==0 part detect that its an even number?

 

[kaliprasad]

x %2 gives the remainder of x when divided by 2.

for odd numebr x %2 = 1 and for even number x % 2 = 0

so this check is made

 

[ineedhelpnow]

will it always be either x%2==0 or x%2==1?

 

[kaliprasad]

will it always be either x%2==0 or x%2==1?

Yes it shall be as when we divide by 2 remainder can be 0 or 1 and no other value

 

[Nono713]

Yes it shall be as when we divide by 2 remainder can be 0 or 1 and no other value

As long as userNum is nonnegative. If it is negative then the result will be 0 or -1, as defined by the C++ standard (which itself follows the Fortran definition of the modulo operation). In short, % is not "modulo", it's "division remainder", which may be negative as integer division always rounds towards zero, not negative infinity.

For instance, let $q$ be the quotient of $-3$ divided by $2$, which is $q = -1$ (round towards zero). Then $r = (-3) ~ \% ~ 2$ is defined as the unique integer satisfying $2q + r = -3$, that is, $r =-1$. So (-3) % 2 = -1.

That is also why I recommend only using % when working with unsigned integer types. It's very easy for people to mistake it for a modulo operation and introduce major security issues that can become very hard to find.

(well, to be specific up until the recent C++11, the sign of (-3) % 2 was implementation-defined, but essentially every processor on the planet will have returned -1; now, it must return -1)​