(no subject)

On Wed, 23 Oct 2002, Nivedita Singhvi wrote:

"Richard B. Johnson" wrote:

quoted

No. It's done over each word (short int) and the actual summation
takes place during the address calculation of the next word. This
gets you a checksum that is practically free.

Yep, sorry, word, not byte. My bad. The cost is in the fact 
that this whole process involves loading each word of the data
stream into a register. Which is why I also used to consider
the checksum cost as negligible. 

quoted

A 400 MHz ix86 CPU will checksum/copy at 685 megabytes per second.
It will copy at 1,549 megabytes per second. Those are megaBYTES!

But then why the difference in the checksum/copy and copy?
Are you saying the checksum is not costing you 864 megabytes
a second??

Costing you 864 megabytes per second?
Lets say the checksum was free. You are then able to INF bytes/per/sec.
So it's costing you INF bytes/per/sec?  No, it's costing you nothing.
If we were not dealing with INF, then 'Cost' is approximately 1/N, not
N. Cost is work_done_without_checksum - work_done_with_checksum. Because
of the low-pass filter pole, these numbers are practically the same.
But, you can get a measurable difference between any two large numbers.
This makes the 'cost' seem high. You need to make it relative to make
any sense, so a 'goodness' can be expressed as a ratio of the cost and
the work having been done.

Cheers,
Dick Johnson
Penguin : Linux version 2.4.18 on an i686 machine (797.90 BogoMips).
   Bush : The Fourth Reich of America