On Mon, Aug 12, 2019 at 01:56:46PM -0400, Lee wrote:
> What's the difference between ${d} and "${d}"? Or is that a bashism
> also? (all my scripts use /bin/sh so I'm pretty clueless wrt bash)
This applies to both sh and bash.
An unquoted substitution, like $d or ${d}, undergoes several steps. The
first step is actually copying the contents of the variable. After that
comes word splitting (dividing the content into words/fields using IFS),
and then pathname expansion ("globbing").
I have a helper script called "args" which I use to illustrate this stuff.
wooledg:~$ cat bin/args
#!/bin/sh
printf "%d args:" "$#"
printf " <%s>" "$@"
echo
Using that, we can demonstrate:
wooledg:~$ d="a variable"
wooledg:~$ args "$d"
1 args: <a variable>
wooledg:~$ args $d
2 args: <a> <variable>
The curly braces don't matter in this case, because there's nothing after
the $d for it to matter.
wooledg:~$ args ${d}
2 args: <a> <variable>
The curly braces are only needed because of the _stuff after the d.
Without them, d_stuff is treated as a variable name.
wooledg:~$ args "$d_stuff"
1 args: <>
wooledg:~$ args "${d}_stuff"
1 args: <a variable_stuff>
The quotes are still needed. Without them, we still get word splitting
and pathname expansion.
wooledg:~$ args ${d}_stuff
2 args: <a> <variable_stuff>
For more details, see <https://mywiki.wooledge.org/Quotes>.
