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
– Ikky, Commented May 16, 2014 at 13:51
hmm, just a througt i got now... putInt is probably Int32? :P — Ikky
– Ikky, Commented May 16, 2014 at 13:52

Sanjeev · Accepted Answer · 2014-05-16 13:54:19Z

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, 2014 at 13:54

Sanjeev

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2 Comments

Yes, thought about that the minute after i asked :/ Thanks

If it helped you, kindly accept as answer :) Thanks

Sotirios Delimanolis · Accepted Answer · 2014-05-16 13:52:46Z

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, 2014 at 13:52

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