Bash bad substitution with subshell and substring

Question

A contrived example... given

FOO=\"/foo/bar/baz\"

this works (in bash)

BAR=$(basename $FOO) # result is BAR=\"baz\"
BAZ=${BAR         


        

Answer 1

It fails because ${BAR:0:1} is a variable expansion. Bash expects to see a variable name after ${, not a value.

I'm not aware of a way to do it in a single expression.