Question

我是目标C的新鲜事。

I need to know how to access the instance variable using the accessors. I could able to access integer variables but not the character variable which i have declared here. Please correct if there is any wrong in this code below.

#define SIZE = 4096

 @interface data : NSObject

  {

    unsigned char id[SIZE];
  }


  @property(readwrite)unsigned char id[SIZE];

  @end

  @implementation data

  @synthesize unsigned char id[SIZE];

  @end

  main.m

  someClass* classPointer = [[someClass alloc]init];

  data* dt = [[data alloc]init];

  [classPointer createMessage:data.id];

Answer 1

为什么不使用<代码>NSString或NSData ,例如,变数而不是阵列? 例如:

@interface Foo2 : NSObject
{
    NSString *_dataId;
}

@property (nonatomic, retain) NSString *dataId;

@end


@implementation Foo2

@synthesize dataId = _dataId;

@end

否则,你就不得不做这样的事情:

#define DATA_ID_SIZE 4096

@interface Foo : NSObject
{
    char _dataID[DATA_ID_SIZE];
}

@property (nonatomic, assign) const char *dataID;

@end

@implementation Foo

// Returns a copy of the internal array of chars.
- (const char *)dataID
{
    size_t length = strlen(_dataID);
    // Dynamically allocate an array of chars to return.
    char *buf = malloc(length);
    // Copy the values from the internal array.
    memcpy(buf, _dataID, length);

    return buf;
}

- (void)setDataID:(const char *)dataID
{
    // Copy provided chars into the internal array.
    memcpy(_dataID, dataID, DATA_ID_SIZE);
    // To be on the safe side, copy null character to the last element.
    _dataID[DATA_ID_SIZE - 1] =   ;
}

这样,<代码>id就成为目标C中的一种数据类型,因此最好不要将它作为变称。

Answer 2

页: 1 因此,它只是 n。

同样,在另一方面,你必须释放你所占领的物体。

Answer 3

如何管理记忆? <代码>C[size]是一个阵列,其大小将按类别成员大小计算,但以同样方式申报的财产将改为/标准pointer,而不是数据! 我建议您使用NSData*(或NSMutableData),而不是C-array,而是Obj-C的首选方式。

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