Explanation of JVM bytecode

Question

Can someone explain how the numbers alongside JVM Opcodes are calculated? I think it is like 1 byte for the opcode and rest of the bytes for operands. Am I correct?

Example:

Method int add12and13() 
 0 bipush 12 
 2 bipush 13 
 4 invokestatic #3 // Method Example.addTwoStatic(II)I 
 7 ireturn

If you are interested in such topics, you should start at the official source first. — Holger
– Holger, Commented Apr 2, 2016 at 13:13
Don’t forget to accept an answer if it satisfies your needs. Otherwise, add to your question what is still unclear. — Holger
– Holger, Commented Apr 8, 2016 at 14:15

apangin · Accepted Answer · 2016-04-01 10:14:08Z

4

You are right. This is bytecode offset from the method beginning.

bipush has 1 byte parameter, so it totally takes 2 bytes.

invokestatic takes 3 bytes: opcode + 2 bytes for a constant pool index, that is, offset of the next instruction will be +3 bytes from this invokestatic.

answered Apr 1, 2016 at 10:14

apangin

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