Bash – i/o format with printf in bash-shell

The code "%$width.${width}s\n" generates a format string that is suitable for consumption by printf

In the script that you posted, width has been assigned the value 55, so that both $width and ${width} are expanded by bash to 55: the entire first parameter to printf expands to %55.55s\n; this is the format %s, with a field width and precision specifiers that ask to print on exactly 55 characters. Given the value of variable divider at this point, this will simply print a line of 55 equal signs. A perhaps simpler way of printing the same thing would have been perl -e 'print "=" x 55, "\n"'.

The simplest form for the field width specifier is an integer: this asks printf to use at least this many characters for printing. If the corresponding parameter requires fewer characters than this to print, then the output is left-padded with spaces.

The simplest form for the precision specifier is a dot followed by an integer: when applied to %s, this sets the maximum number of characters to print. (It has a different meaning for numeric types.)

In response to a comment, I will also mention a little bit about shell variable expansion (the complete explanation can be found by searching for "parameter expansion" in the bash documentation, see also $VAR vs ${VAR} and to quote or not to quote):

• When a variable, say x, has been set, then $x expands to the value of x. If that value contains whitespace, then the expansion will be several words. This is why it's important in the code above that, for example, "$format" is within double quotes: this forces the expansion to be a single word (otherwise, printf would see a first parameter %-15s, followed by an argument %8s, etc., instead of receiving the entire format string as a single parameter).

• It is permissible to write ${x} instead of just $x to expand the variable x; in the case above, "${width}s", it is necessary to do so because if one had written "$widths", then bash would try to get the value of variable widths, which is unset, resulting in an empty expansion.

In fact you need a mutex :

Each sub-shell will access concurrently to the same /dev/stdout of the parent shell, so you can not ensure the order even inside the same function. To ensure it you need a lock which enforce the mutual exclusion i.e. : all other processes won't start to write in /dev/stdout until the lock is released.

#!/bin/bash

function foo {
lockdir=/tmp/myscript.lock
 mkdir "$lockdir" 2>/dev/null
while [ $? -ne 0 ]; do mkdir "$lockdir" 2>/dev/null; done
printf "Test line break: $1\nafter line break: $1\n\n"
rm -rf $lockdir
}

for VARIABLE in {1..30}
do
  foo $VARIABLE &
done
wait

This will give this as a result :

$ bash plop1 2>/dev/null
Test line break: 5
after line break: 5

Test line break: 3
after line break: 3

Test line break: 11
after line break: 11

Test line break: 23
after line break: 23

Test line break: 14
after line break: 14

Test line break: 17
after line break: 17

Test line break: 24
after line break: 24

Test line break: 21
after line break: 21

Test line break: 27
after line break: 27

Test line break: 6
after line break: 6

Test line break: 2
after line break: 2

Test line break: 9
after line break: 9

Test line break: 26
after line break: 26

Test line break: 29
after line break: 29

Test line break: 20
after line break: 20

Test line break: 1
after line break: 1

Test line break: 12
after line break: 12

Test line break: 4
after line break: 4

Test line break: 13
after line break: 13

Test line break: 10
after line break: 10

Test line break: 15
after line break: 15

Test line break: 28
after line break: 28

Test line break: 25
after line break: 25

Test line break: 19
after line break: 19

Test line break: 18
after line break: 18

Test line break: 8
after line break: 8

Test line break: 7
after line break: 7

Test line break: 16
after line break: 16

Test line break: 22
after line break: 22

Test line break: 30
after line break: 30