我试图将一分钟内收到的所有消息对象存储到树形图中,并在一分钟结束后,序列化它并将byte []返回到另一个类,同时清除地图并开始存储下一个接收到的消息分钟等等。
public class StoreMessage extends Thread implements Serializable{
public static byte[] b=null;
public static Map <Long,Message> map1=Collections.synchronizedMap(new TreeMap<Long,Message>());
public static Calendar c1=Calendar.getInstance();
public static int year=c1.get(Calendar.YEAR);
public static int month=c1.get(Calendar.MONTH);
public static int day=c1.get(Calendar.DAY_OF_MONTH);
public static int hour=c1.get(Calendar.HOUR_OF_DAY);
public static int min=c1.get(Calendar.MINUTE);
public static GregorianCalendar gc = new GregorianCalendar(year, month, day, hour, min);
public static Date d=gc.getTime();
public static long time1=(d.getTime())/60000; //precision till minute of the time elapsed since 1 Jan 1970
public static long time2=time1+1;
public static byte[] store(Message message)throws Exception{
while(true)
{
if(time1<time2)
{
long preciseTime=TimeUnit.MILLISECONDS.toNanos(System.currentTimeMillis())+(System.nanoTime()-startNanotime);
map1.put(preciseTime, message);
}
else
{
b=Serializer.serialize(map1);
map1.clear();
time1=time2;
time2++;
return b;
}
}
}
}
为什么这段代码给我空指针异常,突出显示另一个类的 int len = b.length; ,调用它来返回值?
public static void record(Message message){
try{
//storing the serialized message in byte[]
byte[] b =StoreMessage.store(message);
int len=b.length; //<--highlights this line for null pointer exception
即使在进行了修正(即将返回放在else块中)之后,它也不会将控件返回给调用类。此外,在else块中不会打印 SOP 语句(添加时)。为什么呢?
The Serializer class
public class Serializer {
//serializes an object and returns a byte array
public static byte[] serialize(Object map) throws IOException
{
ByteArrayOutputStream b = new ByteArrayOutputStream();
ObjectOutputStream o = new ObjectOutputStream(b);
o.writeObject(map);
return b.toByteArray();
}
//de-serialization of the byte array and returns an object
public static Object toObject (byte[] bytes)
{
Object obj = null;
try
{
ByteArrayInputStream bis = new ByteArrayInputStream (bytes);
ObjectInputStream ois = new ObjectInputStream (bis);
obj = ois.readObject();
}
catch (Exception ex) { }
return obj;
}
}
答案 0 :(得分:0)
这是因为您的b
变量为null
。见这里:
您输入方法并进行检查if(time1<time2)
。这是true
。因此你不进入其他并且不初始化b。之后你会去返回b
的值。它是null
。如果你问我,你已经错误地放置了返回声明。将返回值放在else语句中,这样就可以确保循环终止,并且在返回之前仍然会初始化数组:
while(true) {
if(time1<time2) {
long preciseTime=TimeUnit.MILLISECONDS.toNanos(System.currentTimeMillis())+(System.nanoTime()-startNanotime);
map1.put(preciseTime, message);
} else {
b=Serializer.serialize(map1);
map1.clear();
time1=time2;
time2++;
return b;
}
}
}