将UITouch保存到NSDictionary

Question

首先，我尝试归档一个包含UITouch的NSDictionary。我一直收到这样的错误：“-UITouch encodeWithCoder:：unrecognized selector sent to instance”。

如果从字典中删除UITouch对象，则不会收到错误。

我自己一直在尝试解决这个问题，包括在过去的几个小时里搜索谷歌，但还没有找到如何在NSDictionary中存储UITouch对象的方法。

下面是我使用的方法：

- (void)sendMoveWithTouch:(id)touch andTouchesType:(TouchesType)touchesType {
     MessageMove message;
     message.message.messageType = kMessageTypeMove;
     message.touchesType = touchesType;
     NSData *messageData = [NSData dataWithBytes:&message length:sizeof(MessageMove)];

     NSMutableData *data = [[NSMutableData alloc] init];
     NSKeyedArchiver *archiver = [[NSKeyedArchiver alloc] initForWritingWithMutableData:data];

     [archiver encodeObject:@{@"message" : messageData, @"touch": touch} forKey:@"touchesDict"]; //RECEIVE ERROR HERE. If I remove the UITouch object, everything passes correctly

     [archiver finishEncoding];

     [self sendData:data];
}

任何帮助都将不胜感激。

Answer 1

UITouch本身并不符合<NSCoding>协议，因为它没有简单/明显的序列化表示(比如字符串或数组，它们是基本的数据类型)。你要做的就是通过扩展它的类来使它符合这个协议，并决定它应该序列化哪些属性以及以何种格式序列化。例如：

@implementation UITouch (Serializable)

- (void)encodeWithCoder:(NSCoder *)coder
{
    [coder encodeObject:@([self locationInView:self.view].x) forKey:@"locationX"];
    [coder encodeObject:@([self locationInView:self.view].y) forKey:@"locationY"];
}

- (id)initWithCoder:(NSCoder *)decoder
{
    if (self = [super init]) {
        // actually, I couldn't come up with anything useful here
        // UITouch doesn't have any properties that could be set
        // to a default value in a meaningful way
        // there's a reason UITouch doesn't conform to NSCoding...
        // probably you should redesign your code!
    }
    return self;
}

@end