How to calculate input (value &; Key) in cache size?
Guys,
I see two different values when you try to calculate the size of an entry (key & value) object in the cache, you wonder which is correct
I <BINARY>Unit-Calculator< /Unit-Calculator> and untiFactor = 1
1. with the help of ExternalizableHelper
Serializer cache serializer = .getCacheService () .getSerializer ();
int entrySize = ExternalizableHelper. toBinary (tKey, serializer). length() + ExternalizableHelper. toBinary(tKey, serializer). length();
output entrySize = 130bytes
2 mark MBean JMX
< ObjectName > game names = mbeanServer.queryNames (new ObjectName ("consistency: type = Cache, service = *, name =" + cacheName + ", *"), null);
long totalUnits = 0;
for (name ObjectName: name) {}
Units of Integer = (integer) mbs.getAttribute (name, 'Units');
Integer unitFactor = (Integer) mbs.getAttribute (name, "UnitFactor");
Size of the integer = (units * unitFactor) / size;
}
output size = 240bytes...
So, ask yourself why there is a huge difference in the size returned by ExternalizableHelper and JMX Bean.
1. don't JMX MBean also add coherence an overload to the size of the entry?
2. which is the correct method for compensable size entry?
Thank you
D
Hello
Yes the BinaryCalculator is given overhead.
JK
Tags: Fusion Middleware
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