Question: When I construct a new `BufferedReader` it is providing me a buffer of 8192 characters. What is the logic/reason behind this?

```
8192 = 2 to the power of 13

```

Answer: 
Traditionally, memory managers and paging files in the operating system work on pages that are sized in powers of 2. This allows very efficient multiply/divide operations to be performed with left/right shift operations. When working with a buffer, the worst case scenario is to have a buffer with size 1 byte longer than the page size (that would result in an extra page swap with very low benefit). So the default buffer sizes will also tend to be implemented in factors of two.

I'd assume (but have not checked) that the JVM looks for buffers like this and attempts to align them on page boundaries.

Why does this matter? Page misses are quite expensive. If you are doing a ton of IO, it's better to avoid the case where the page backing the buffer gets swapped out to disk (kind of defeats the purpose of the buffer). That said, for most applications, this is a micro-optimization, and for the vast majority of cases, the default is fine.

For reference, Windows and Linux both currently use a 4KB memory page size. So the default buffer on BufferedReader will consume exactly 2 pages.
8192 is 2^13 and also reveals much information regarding the RIGHT v WRONG encoded in all we do. If one takes away or adds to the author intent, he modifies and therefore corrupts the entire thing. Try to add or take away from something perfect... good luck!