I've come across this solution for printing a specific line from a text file:

sed '123!d;q' file

Why doesn't sed quit after the first line of input in this case?

1 Answer 1


In English, this sed program means: for each line,

  • [123!] if the current line number is not 123, then [d] delete the current line and start the next cycle (i.e. move to the next line);
  • then (but we only reach this point if the d command was not executed), [q] exit without processing any more lines (but do print out the current line in our dying throes).

Or if you prefer, in shell syntax:

while IFS= read -r pattern_space; do
  if [ $line_number -ne 123 ]; then       # 123!
    continue                              #   d
  echo "$pattern_space"; break            # q
  echo "$pattern_space"                   # implicit final print (never reached)
  • Thanks, that cleared it to me also. My error was that I understood ! was connected to d, not 123. Nov 18, 2011 at 8:44
  • @Gilles: missing the braces {d;q;}, shouldn't q apply to every line (so only the first)?
    – enzotib
    Nov 18, 2011 at 9:31
  • 1
    @enzotib q applies to every line where it's executed. But when the line number is not 123, the d command is executed, and its meaning is to skip immediately to the next input line. Nov 18, 2011 at 10:18
  • Crystal clear explanation Jan 24, 2014 at 15:44
  • Essentially it does the same as sed -n 123p (prints the same output) except it stops after line 123 rather than processing potentially thousands more lines that it will never do anything with anyways.
    – Wildcard
    Feb 15, 2016 at 7:33

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