Issue
I am trying to set up a variable that contains a string representation of a value with leading zeroes. I know I can printf to terminal the value, and I can pass the string output of printf to a variable. It seems however that assigning the value string to a new variable reinterprets the value and if I then print it, the value has lost its leading zeroes.
How do we work with variables in bash scripts to avoid implicit type inferences and ultimately how do I get to the solution I'm looking for. FYI I'm looking to concatenate a large fixed length string numeric, something like a part number, and build it from smaller prepared strings.
Update:
Turns out exactly how variables are assigned changes their interpretation in some way, see below:
Example:
#!/bin/bash
a=3
b=4
aStr=$(printf %03d $a)
bStr=$(printf %03d $b)
echo $aStr$bStr
output
$ ./test.sh
003004
$
Alternate form:
#!/bin/bash
((a = 3))
((b = 4))
((aStr = $(printf %03d $a)))
((bStr = $(printf %03d $b)))
echo $aStr$bStr
output
$ ./test.sh
34
$
Solution
By doing ((aStr = $(printf %03d $a))), you are destroying again the careful formatting done by printf. You would see the same effect if you do a
(( x = 005 ))
echo $x
which outputs 5.
Actually the zeroes inserted by printf could do harm to your number, as you see by the following example:
(( x = 015 ))
echo $x
which outputs 13, because the ((....)) interprets the leading zero as an indication for octal numbers.
Hence, if you have a string representing a formatted (pretty-printed) number, don't use this string in numeric context anymore.
Answered By - user1934428 Answer Checked By - Clifford M. (WPSolving Volunteer)
