Hi,

David Alan Black wrote:


>   [1,1].minus([1,1,1]) => [1]
> 
> ?
> 
> If so, you'd have to differentiate the arrays based on size:


that's what I did first.

But there are problems like [1,2].diff [3,4,5]; here, [1,2] should be returned, not [3,4,5].


The first method I wrote was named diff, and acted accoring to the size; 
but then the above problem occured.

So something like #minus is better and way clearer; one gets subtracted 
from the other. It's up to the programmer to detect the size.


Tobi


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