[SOLVED] fast BASH script question

pan64 · 02-03-2019, 04:07 AM

Quote:

Originally Posted by lleb

Code:

for i in {0..6..1}	# {START...END...INCREMENT}
	do
		# echo "foo.${i}.TEST"&>>grab.txt
		printf 'foo.%02d'${i}'.TEST\n'&>>grab.txt
	done

Question though, why %02d and not %03d.

In this example you print a 2 digit number and $i. printf requires a format string and arguments. Here the argument is missing, so it is 0 always. That's why you have 00 always before $i.

From the other hand seq executed in a subshell, therefore it is much slower than any other solution.

lleb · 02-03-2019, 06:14 PM

Quote:

Originally Posted by Beryllos

Small correction about your comments:

in the format specifier '%03d'
d means the number is expressed in decimal (base 10)
3 is the minimum field width, normally whitespace-padded
0 signifies zero padding

Code:

$ printf '%d\n' "5"
5
$ printf '%3d\n' "5"
  5
$ printf '%03d\n' "5"
005
$

Thank you for the details on the comments. I have updated my code to make it simpler to repeat, edit down the line.