CODEWITHSUNDEEP
c#if-statementstatements
EssentialAnonymity
EssentialAnonymity

Reputation: 316

Can you explain this code for me in layman's terms?

I'm doing a project where I need the code to be able to determine if input is odd or not. (I also need to be able to determine if it is even

Can you explain what this means? Mainly the first part.

if((x & 1) == 0)
    printf("EVEN!\n");
else
    printf("ODD!\n");

Upvotes: 2

Views: 217

Answers (3)

Dan
Dan

Reputation: 1080

It's a bitwise operation. The AND operation masks every bit and leave the last one. If the last bit is off (you're checking against zero), means it's even otherwise means it's odd.

An example:

125 in binary form is 01111101

so

01111101 AND
00000001 =
------------
00000001

so 125 is ODD :)

An alternative way to check even or odd is to use the module of division:

if (x % 2 == 0) printf("EVEN");
else printf("ODD");

As noted by a user in a comment below, this second solution is slower (in terms of μS)

Upvotes: 8

Derek
Derek

Reputation: 8753

A simpler way to do it would be to:

if( x % 2 == 0 )
{
  printf("EVEN!\n");
}
else
{
   printf("ODD!\n");
}

The %, or modulus operator, returns the remainder.

http://msdn.microsoft.com/en-us/library/0w4e0fzs.aspx

Upvotes: 0

ngneema
ngneema

Reputation: 444

As stated in the other answer it is a bitwise operation, but the test is actually if the last bit is set it is ODD, hence the ==0. So if the last bit is NOT set it is EVEN.

EDIT: Just saw that answer was deleted, oh well.

Upvotes: 2

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