hi,

Just came across someone using sed with the following way:

which will eventually print the line no.10 of the logfile. I have never thought it can be use that way. However I don't understand how it works.

I understand by using will print the first 10 line and simulate the function of ''head''. But what does the ''d'' do that delete the previous 9 line ? What logic behind ?

I then tried with , it prints the 5th line of logfile. How does it work this time ?


Anyone would like to share ? Just for knowledge purpose, nothing specific function to achieve here.

Thanks for sharing though

Well, you have to think at the two commands executed in sequence on every line of the input file: read first line, don't quit, delete, read second line, don't quit, delete.... read tenth line, quit. This means the first nine lines are deleted and when the 10th line is reached the program terminates and the line is printed out.

The same for the second example: every line before the 5th one is deleted and when the 5th line is reached the program terminates and it's printed out. The second quit command (45q) is irrelevant, since that line is never reached.

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Hi.
This behaviour becomes pretty straightforward when you remember that sed reads each line into a buffer called pattern space, evaluates all given commands and, if no `-n' option was specified, prints its contents, in this order. So, for lines 1-9, it reads a line, then evaluates `10q', which has no effect on these lines, then evaluates `d', which clears pattern space, then it tries to print pattern space, which is empty, so nothing is printed. The same happens for lines 11, 12, etc. For line #10, `10q' makes sed to quit immediately (skipping all other commands), but before it quits the pattern space is printed.

PS: I'm late again.

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It's ok, your reply does give more input to me about the pattern space logic, appreciated that too

With colucix explanation, I tried just now:
, it gives empty result, which means the d delete the pattern space before it even reaches to 5q. So using with the extra command behind d will give no meaning as the d in front already delete everything in the pattern space. Correct ?

And if we use , the q will only processed at the 5th line and print it out, other lines below will not be processed so it make no sense to put more q inside a sed command. Am I right in understanding this ?

This q comes out to be a cool way in printing one particular line from a pretty large file. I tested just now compare with p and d: and , the q way really give faster result with a huge file.

Yes, all correct.

Thanks firstfire for the in-depth explanation!