Why do bytebuffer give buffer overflow exception when buffer is not full

Question

I'm not sure why the following example gives buffer overflow exception. Hope someone can explain why, and how i can do it correctly.

It's as simple as this:

ByteBuffer bf = ByteBuffer.allocate(4);
bf.order(ByteOrder.BIG_ENDIAN);
bf.putInt(8);
bf.putInt(7); // Throws exception

The goal: [0,0,8,7]

Thanks in advance!

haha, sorry. Forgot to insert a number instead of the variable. 4 — Ikky, May 16 '14 at 13:51
hmm, just a througt i got now... putInt is probably Int32? :P — Ikky, May 16 '14 at 13:52

score 7 · Accepted Answer · answered May 16 '14 at 13:54

7

An int is 4 bytes long so you should multiply 4 to the number of int you need to store in your ByteBuffer.

answered May 16 '14 at 13:54

Sanjeev

Yes, thought about that the minute after i asked :/ Thanks – Ikky May 16 '14 at 13:55
If it helped you, kindly accept as answer :) Thanks – Sanjeev May 16 '14 at 14:54

score 1 · Answer 2 · answered May 16 '14 at 13:52

1

The javadoc states

BufferOverflowException - If there are fewer than four bytes remaining in this buffer

Your totalNumberOfBytes must not be big enough to fit 2 ints, ie. less than 8.

answered May 16 '14 at 13:52

2 Answers2