From the manpage:
...so you may be corrupting your clock() call by running commands in system(), though it's still odd that it'd mung it that far. Try removing the system() calls and just printing the clock() values, running the program with 'time ./program' to see how your values compare to the ones time sees.
Hey ppl, i was wonddering, in mandrake, how to get the clok to display the time in non-military format....hehe thank you im just tired of looking at 18:00 hehe thank you (2 Replies)
Guys could you please tell me which appropriate command is used to set hardware (BIOS) clock so that the system keeps time when it reboots & how it's used. I use Linux
Thank you (2 Replies)
Hey all,
i need a program to get the CPU ticks at certain points of my program. So, i thought about using the clock function, but i'm having a hard time figuring out how it really works. I wrote this simple program to try to understand it but it made me feel more confused:
#include <stdio.h>... (5 Replies)
Hi,
Is there a chance that the clock() call returns 0 eternally???
Using BSD. My RTOS application freezes inconsistently only on particular hosts. When debugging it, I came to see that the RTOS timer does not tick at times. The underlying system call is clock() & it always returns zero when the... (4 Replies)
Hi
We had a AIX box built last year but was set to the correct GMT time, but using DST time zone. In march this year the clocks went forward without issues. (if I remember a couple of weeks early due to the DST zone)
This year we decided to change the clock to the correct time zone before... (0 Replies)
Hi there!!!
Need your help in solving some tricky problems.
Since clock() as such is buggy on SUN OS 5 we have started using gettimeofday() in our RTOS applications based on Solaris 9.
The problems we actually encountered previously were - the applications kind of freeze/hang eternally on... (1 Reply)
Hi all
Hi could anyone tell me how i can set the
Hardware Clock to the System Time, and set the System Time from the
Hardware Clock.
i am using RHEL 4.0.
Thanks in advance. (1 Reply)
I am trying to implement a simple chess clock. It should have the following options: start, stop, reset, read.
Reset will set the time to zero
Start will start the clock
Stop will stop the clock
My problem is that I want that start continues counting
the time from the time it had when it... (6 Replies)
Discussion started by: kristinu
6 Replies
LEARN ABOUT DEBIAN
bup-margin
bup-margin(1) General Commands Manual bup-margin(1)NAME
bup-margin - figure out your deduplication safety margin
SYNOPSIS
bup margin [options...]
DESCRIPTION
bup margin iterates through all objects in your bup repository, calculating the largest number of prefix bits shared between any two
entries. This number, n, identifies the longest subset of SHA-1 you could use and still encounter a collision between your object ids.
For example, one system that was tested had a collection of 11 million objects (70 GB), and bup margin returned 45. That means a 46-bit
hash would be sufficient to avoid all collisions among that set of objects; each object in that repository could be uniquely identified by
its first 46 bits.
The number of bits needed seems to increase by about 1 or 2 for every doubling of the number of objects. Since SHA-1 hashes have 160 bits,
that leaves 115 bits of margin. Of course, because SHA-1 hashes are essentially random, it's theoretically possible to use many more bits
with far fewer objects.
If you're paranoid about the possibility of SHA-1 collisions, you can monitor your repository by running bup margin occasionally to see if
you're getting dangerously close to 160 bits.
OPTIONS --predict
Guess the offset into each index file where a particular object will appear, and report the maximum deviation of the correct answer
from the guess. This is potentially useful for tuning an interpolation search algorithm.
--ignore-midx
don't use .midx files, use only .idx files. This is only really useful when used with --predict.
EXAMPLE
$ bup margin
Reading indexes: 100.00% (1612581/1612581), done.
40
40 matching prefix bits
1.94 bits per doubling
120 bits (61.86 doublings) remaining
4.19338e+18 times larger is possible
Everyone on earth could have 625878182 data sets
like yours, all in one repository, and we would
expect 1 object collision.
$ bup margin --predict
PackIdxList: using 1 index.
Reading indexes: 100.00% (1612581/1612581), done.
915 of 1612581 (0.057%)
SEE ALSO bup-midx(1), bup-save(1)BUP
Part of the bup(1) suite.
AUTHORS
Avery Pennarun <apenwarr@gmail.com>.
Bup unknown-bup-margin(1)