UNIX for Dummies Questions & Answers

Hi Guys,

i'm exploring sed and failed to understand the following. Can anyone with more knowledge of this explain this better.

I have to read lines 4 to 6 in a file so i used the following command :

Code:

sed '4,6 p' file

but the above prints all lines instead! . when i use the -n switch however it prints the desired lines only.

I know that the -n switch prevents sed from printing to console but how exactly is it affecting my above example. Please advise.

Thanks

From the sed man page...

Code:

-n    Suppress the default output (in which each line, after it is examined for editing, is written to standard output). Only lines explicitly selected for output are written.

thanks guys. this explains it..i'm using cygwin on my home laptop and the man page on there was not that explicit.

Cheers

A little more explanation....

sed works line by line. Each line in the input is read into a pattern space and then all the edits in your script are attempted on that pattern space. At the end of the script, the pattern space is output, irrespective of whether any or all or none of the edits affected the pattern space. This is the default behaviour of sed. That is precisely why it's called a stream editor/filter. It works on a stream; what goes in, comes out, albeit changed in some cases.

The -n switch (or the #n comment line in scripts) suppresses this default behaviour. In such cases, sed will output the pattern space only when it is told to do so explicitly.

In your command (without the -n switch), lines 1-3 and those after line 6 will just pass through the script unaffected. Lines 4-6 will match the address of the print command and will be output:
1) by your print command, and then
2) by sed's default behaviour.