I accidentaly wrote:
printf "here: %s $d\n" "${line:0:32}" "${#line}"
and got:
here: /home/gsamaras/Bash/libs/goodLib
here: 103
Why?
Of course I meant to say %d, but I don't understand this behavior in the mistake I made. I would probably expect it to print "here: /home/gsamaras/Bash/libs/goodLib $d", like it would do in C... I couldn't find a duplicate or something on that, thus the question.
Before the string "here: %s $d\n" is passed to printf, parameter expansions are performed by the shell. In this case $d is expanded to an empty string.
If you used single quotes around the string, or backslash-escaped the $, then you would see $d in the output.
Since you only have one format specifier in your string (%s) and passed two arguments after the format string, you end up with two lines of output:
The format is reused as necessary to consume all of the arguments.
(from man bash in the printf section).
First step, the shell performs variable expansion. Since there's no variable $d, it's replaced with an empty string. After variable replacement it's as if you'd written:
printf 'here: %s \n' /home/gsamaras/Bash/libs/goodLib 103
Now why does it print here: twice? When printf is given more arguments than format specifiers it repeats the format string, looping extra times until it's consumed all of its arguments. Since you have a single %s but two extra arguments it loops twice. It's as if you had written:
printf 'here: %s \n' /home/gsamaras/Bash/libs/goodLib
printf 'here: %s \n' 103
That is why you get two lines of output.
