John Williams wrote: > Back in October I suggested that $a ^+= @b would act like reduce, > but in discussion > it was decided that it would act like length
> I now pose the question: Is ^+= a "hyper assignment operator" or an > "assignment hyper operator"? > with a scalar involved > the method and the result is different. $a = length(@b) is the > "assignment hyper operator" > interpretation. The "hyper assignment operator" interpretation looks > like this: > > $a ^+= @b > ($a, $a, $a, ...) ^+= @b > $a += $b[0], $a += $b[1], $a += $b[2], ... I can't remember what side I argued last October (I can't remember last October! %-) I have to say that I'm with John here. That interpretation certainly seems more DWIM to me. Damian
