Checking if a NSInteger is odd or even.

Question

I have been trying to check whether an NSInteger is odd or even. I have found a way to do it using C but it doesn't work with Objective-C. How would I do this?

Answer 1

NSInteger n = 5;

NSLog(@"%s", n & 1 ? "odd" : "even");

or using if

if (n & 1) {
  ; // odd
} else {
  ; // even
}

with some output:

if (n & 1) {
  NSLog(@"odd");
} else {
  NSLog(@"even");
}

the pointer example:

NSInteger x = 7;
NSInteger *y = &x;

if (*y & 1) {
    NSLog(@"odd");
} else {
    NSLog(@"even");
}

Answer 2

NSInteger is defined as int (or long on some environments). So checking on oddity is like for plain int:

NSInteger num;
if (num % 2)
  // odd
else
  // even

Answer 3

As far as I'm aware. NSInteger, unlike NSNumber, is just a typeder to a real integer type along the lines of:

typedef long NSInteger;

So you should be able to do:

NSInteger nsintvar = 77;
if ((nsintvar % 2) == 0) {
    // number is even
} else {
    // number is odd
}

---

Here's a complete program, compiled under Cygwin with GNUstep, which illustrates it:

#import <stdio.h>
#import <Foundation/NSObject.h>

int main( int argc, const char *argv[] ) {
    NSInteger num;
    for (num = 0; num < 20; num++) {
        if ((num % 2) == 0) {
            printf ("%d is even\n", num);
        } else {
            printf ("%d is odd\n", num);
        }
    }
    return 0;
}

It outputs:

0 is even
1 is odd
2 is even
3 is odd
4 is even
5 is odd
6 is even
7 is odd
8 is even
9 is odd
10 is even
11 is odd
12 is even
13 is odd
14 is even
15 is odd
16 is even
17 is odd
18 is even
19 is odd

Answer 4

Those other answers should work. Maybe it's a problem with your makefile or something. Think outside that piece of code.

If all else fails just declare the integer as an int. You don't have to declare it as NSInteger.