What does $d do in printf?

Question

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.

Answer 1

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).