在 Objective-C 中传递指针到函数

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。

下面是一个简单的示例，我们将一个无符号长指针传递给函数并更改函数内部的值，该值反映在调用函数中 -

#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;
}

当上面的代码被编译并执行时，它会产生以下结果 -

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

该函数可以接受指针，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;
}

当上面的代码一起编译并执行时，会产生以下结果 -

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