Understanding regex behaviour when using quantifiers
So, in the above code , sed replaces at the start. does that mean sed using the pattern e* settles to zero occurence ? Why sed was not able to replace Teest string.
I am having trouble parsing rpm filenames in a shell script.. I found a snippet of perl code that will perform the task but I really don't have time to rewrite the entire script in perl. I cannot for the life of me convert this code into something sed-friendly:
if ($rpm =~ /(*)-(*)-(*)\.(.*)/)... (1 Reply)
I have these two files in current dir:
oos.txt
oos_(copy).txt
I execute this find command:find . -regex './oos*.txt'And this outputs only the first file (oos.txt)! :confused:
Only if I add another asterisk to the find find . -regex './oos*.*txt' do I also get the second file... (7 Replies)
Hi,
Please help me to understand the bold segments in the below regex.
Both are of same type whose meaning I am looking for.
find . \( -iregex './\{6,10\}./src' \) -type d -maxdepth 2
Output:
./20111210.0/src
In continuation to above:
sed -e 's|./\(*.\{1,3\}\).*|\1|g'
Output: ... (4 Replies)
I have the following line of code that works wonders. I just don't completely understand it as I am just starting to learn regex. Can you help me understand exactly what is happening here?
find . -type f | grep -v '^\.$' | sed 's!\.\/!!' (4 Replies)
Hi Guys,
Could you please kindly explain what exactly the below SED command will do ?
I am quite confused and i assumed that,
sed 's/*$/ /'
1. It will remove tab and extra spaces .. with single space.
The issue is if it is removing tab then it should be Î right ..
please assist.... (3 Replies)
I am not a big expert in regex and have just little understanding of that language.
Could you help me to understand the regular Perl expression:
^(?!if\b|else\b|while\b|)(?:+?\s+){1,6}(+\s*)\(*\) *?(?:^*;?+){0,10}\{
------
This is regex to select functions from a C/C++ source and defined in... (2 Replies)
Hi everyone,
This regex looks simple and yet it doesn't make sense how it's manipulating the output.
ifconfig -a
eth0 Link encap:Ethernet HWaddr 00:0c:49:c2:35:6v
inet addr:192.16.1.1 Bcast:192.168.226.255 Mask:255.255.255.0
inet6 addr:... (2 Replies)
I'm trying to get some exclusions into our sendmail regular expression for the K command. The following configuration & regex works:
LOCAL_CONFIG
#
Kcheckaddress regex -a@MATCH
+<@+?\.++?\.(us|info|to|br|bid|cn|ru)
LOCAL_RULESETS
SLocal_check_mail
# check address against various regex... (0 Replies)
Hi,
I need some guidance with understanding this Perl script below. I am not the author of the script and the author has not leave any documentation. I supposed it is meant to be 'easy' if you're a Perl or regex guru. I am having problem understanding what regex to use :confused: The script does... (3 Replies)
Hello All,
While googling on regex I came across a site named Regulex Regulex:JavaScript Regular Expression Visualizer
I have written a simple regex ^(a|b|c)(*)@(.*) and could see its visualization; one could export it too, following is the screen shot.
... (3 Replies)
Discussion started by: RavinderSingh13
3 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)