在Objective-C中传递指针给函数

Objective-C编程语言允许您将指针传递给函数。为此，只需将函数参数声明为指针类型即可。

以下是一个简单的示例，我们将一个无符号长整型指针传递给一个函数，并在函数内部更改其值，该值会反映回调用函数中：

#import <Foundation/Foundation.h>
 
@interface SampleClass:NSObject
- (void) getSeconds:(int *)par;
@end

@implementation SampleClass

- (void) getSeconds:(int *)par {
 /* get the current number of seconds */
   *par = time( NULL );
   return;
}

@end

int main () {
   int sec;

   SampleClass *sampleClass = [[SampleClass alloc]init];
   [sampleClass getSeconds:&sec];

   /* print the actual value */
   NSLog(@"Number of seconds: %d\n", sec );

   return 0;
}

编译并执行上述代码后，将产生以下结果：

2013-09-13 23:50:47.572 demo[319] Number of seconds: 1379141447

可以接受指针的函数也可以接受数组，如下例所示：

#import <Foundation/Foundation.h>
 
@interface SampleClass:NSObject
/* function declaration */
- (double) getAverage:(int *)arr ofSize:(int) size;
@end

@implementation SampleClass

- (double) getAverage:(int *)arr ofSize:(int) size {
   int    i, sum = 0;       
   double avg;

   for (i = 0; i < size; ++i) {
      sum += arr[i];
   }

   avg = (double)sum / size;
   return avg;
}

@end

int main () {

   /* an int array with 5 elements */
   int balance[5] = {1000, 2, 3, 17, 50};
   double avg;

   SampleClass *sampleClass = [[SampleClass alloc]init];
   /* pass pointer to the array as an argument */
   avg = [sampleClass getAverage: balance ofSize: 5 ] ;

   /* output the returned value  */
   NSLog(@"Average value is: %f\n", avg );

   return 0;
}

编译并执行上述代码后，将产生以下结果：

2013-09-14 00:02:21.910 demo[9641] Average value is: 214.400000