Shell syntax highlighting

Question

I've noticed in bash/shell it highlights $(...) in red, for example:

However, if I change the syntax from ft=sh to ft=bash, it does not do this:

What's the reason for this? Am I doing something wrong with the ft=sh, or why is that highlighted in red when it's a legitimate syntax construct?

muru · Accepted Answer

This is because the original Bourne shell only understands `...` - the standard $( ... ) syntax was added later. From syntax/sh.vim:
" $() and $(()): {{{1
" $(..) is not supported by sh (Bourne shell).  However, apparently
" some systems (HP?) have as their /bin/sh a (link to) Korn shell
" (ie. Posix compliant shell).  /bin/ksh should work for those
" systems too, however, so the following syntax will flag $(..) as
" an Error under /bin/sh.  By consensus of vimdev'ers!

You can set the variable g:is_bash to tell the sh syntax file that sh is actually bash, if indeed it is actually bash (does sh --version show the Bash version?). Or you can set g:is_posix if sh is POSIX compliant.

D. Ben Knoble · Answer

Shell originally used only backticks for command substitution:
less `find . -type f`

But these do not nest and suffer from other problems. Bash (and probably ksh, but I’m not up on the history there) used $() as a newer, improved command substitution.
But that explains the difference and while the filetype matters—in fact, the shebang (#! line) matters! If you use /bin/sh, you should only use POSIX sh features. Otherwise, use the path to a known interpreter and use it’s features.
As pointed out in the comments, POSIX allows the newer command substitution, so it should be safe even in sh scripts.

Shell syntax highlighting

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