检查NSInteger是奇数还是偶数

Question

我一直试图检查NSInteger是奇数还是偶数.我找到了一种使用C来实现它的方法,但它不适用于Objective-C.我该怎么做？

Answer 1

NSInteger定义为int(或long在某些环境中).所以检查奇怪就像普通的int:

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

Answer 2

NSInteger n = 5;

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

或使用if

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

有一些输出:

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

指针示例:

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

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

Answer 3

据我所知.与之NSInteger不同的是NSNumber,它只是一个实数整数类型的打字机:

typedef long NSInteger;

所以你应该能够做到:

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

这是一个完整的程序,在Cygwin下使用GNUstep编译,它说明了:

#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;
}

它输出:

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