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Operating Systems Solaris Explain @(#)cshrc 1.11 89/11/29 SMI Post 302843471 by fgrirx on Tuesday 13th of August 2013 12:09:32 PM
Old 08-13-2013
Explain @(#)cshrc 1.11 89/11/29 SMI

What does ' @(#)cshrc 1.11 89/11/29 SMI ' mean?
Can someone please deconstruct and explain the parts?
Code:
# @(#)cshrc 1.11 89/11/29 SMI

It is at the top of a the .cshrc of a new Solaris account I am working on.
I am familiar with using the first line of a script for setting the program/shell to use. Example: #!/bin/csh or #!/bin/sh or #!/ben/perl.
I am also familiar with leaving the first line blank or adding a comment. For .cshrc (opposed to my_script.csh) I just start with a # comment line.

Searching the internets, I see lots of examples of .cshrc posted that begin with this line, but I have not found anybody who explains what it is doing or where it comes from. The best I tell, it is just a comment. But why does it appear at the beginning of so many .cshrc files?

Last edited by fgrirx; 08-13-2013 at 07:02 PM.. Reason: bad grammer and misspelling
 
Test Your Knowledge in Computers #535
Difficulty: Medium
In C, the Boolean type false is associated with zero while any non-zero value is associated with true.
True or False?
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)

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