Re: status on $[arith] for eval arith vsl $((arith))??

[Greg Wooledge]

On Wed, Apr 11, 2012 at 02:34:01AM -0700, Linda Walsh wrote:
> - for ((vl=$((v_level - 1)); $vl > 0; --vl))

The inside of the for ((...)) is already a math context.  You don't need
another $((...)) inside it.

for ((vl = v_level - 1; vl > 0; --vl))

Or is that another "irrelevant detail"?