On Wed, Sep 02, 2015 at 11:24:42AM -0400, Chet Ramey wrote: > On 9/2/15 11:19 AM, Greg Wooledge wrote: > > On Wed, Sep 02, 2015 at 10:16:14AM -0500, Dennis Williamson wrote: > >> The $ is implied. > > > > That is completely absurd. (And wrong.) > > Not exactly. When the arithmetic evaluator encounters a token that is of > the form of a shell identifier (`bar'), in a context where an operand is > needed, it treats it as a shell variable and looks up the variable's > value. In that sense, it's an expansion. > > The difference between bash and dash is what each shell does with that > value.
$foo and foo are not equivalent in dash, as we've already discussed: $ dash $ foo=bar bar=5 $ echo $((foo)) dash: 4: Illegal number: bar $ echo $(($foo)) 5 They are also not equivalent in bash, if I may be permitted to change the context slightly: $ bash $ foo='$(id >&2)' $ echo $((x[foo])) bash: $(id >&2): syntax error: operand expected (error token is "$(id >&2)") $ echo $((x[$foo])) uid=563(wooledg) gid=22(pgmr) groups=1002(webauth),208(opgmr) 0 So "The $ is implied" is false, in both shells, although there are some contexts where they happen to produce the same result.
