[physfs] question about PHYSFS_read and PHYSFS_write

[Ryan C. Gordon]

> In some drivers, they are identical (it'll read size*count bytes in one
> block), others will be (theoretically) more efficient to read 8192 bytes
> in one block, but likely do the exact same thing (size*count) under the
> hood.

Some more thoughts:

...the _real_ tradeoff is that if you can't read that whole block, you
get nothing back if you've asked for size=8192 count=1, whereas size=1
count=8192 will return however many bytes were actually available.

The smartest move, really, is to use the physfs buffering API and then
make the read call in whatever way works best for your program. If you do
it right, then most read calls will run at the about the speed of a
memcpy() of size*count bytes and the total i/o overhead reduces the more
you can safely buffer (that is, without running into disk swapping) at
once.

Finally, don't ever do i/o, physfs or otherwise, in time-critical loops. I
assume this is a discussion about lowering startup or mapchange times or
whatnot, but I figured I should throw that out there anyhow.

--ryan.