I would like to print the character at a given position using only the command line. E.g.:

<command> 5

Would output a if the 5th char of that file was a.

Since I am dealing with big files, ideally this would be able to handle big files.


With sed:

$ echo 12345 | sed 's/.\{4\}\(.\).*/\1/;q'
$ echo 1234ắ | sed 's/.\{4\}\(.\).*/\1/;q'

Note that sed will fail to produce output if you input contain invalid multi-byte characters in current locale. You can use LC_ALL=C if you work with single byte characters only.

With ASCII file, you can also use dd:

$ echo 12345 | dd bs=1 skip=4 count=1 2>/dev/null
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    Is there any solution that doesn't involve having to pipe the input? Otherwise it will be slow in bit files no? – testTester Apr 16 '16 at 9:00
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    @testTester: Just use your file as operand to the command sed 's/.\{4\}\(.\).*/\1/;q' file – cuonglm Apr 16 '16 at 9:01

If you want the 5th byte, counting from 1:

dd ibs=1 skip=4 count=1


tail -c +5 | head -c 1

Note that tail counts from 1, so given a file containing abcdefg, this prints e.

dd and tail -c are in POSIX. head -c is common, but isn't in POSIX; it's in GNU coreutils, BusyBox, FreeBSD and NetBSd and but not in OpenBSD or Solaris.

  • It's probably worth mentioning that in their current form, these commands won't do anything. You'll either need to add a filename parameter on to the commands or pipe input into them. e.g.: cat file | tail -c +5 | head -c 1 – rinogo Apr 6 '18 at 15:14
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    @rinogo Both commands read from standard input and write to standard output. – Gilles Apr 6 '18 at 17:08
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    Indeed! My aim was to help those new to Unix understand how to get data into the commands. – rinogo Apr 6 '18 at 17:10

Or using (gnu)grep:

grep -zoP '.{4}\K.'   file

(-z was used to deal with \n before the 5th char)

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