Re: [sv-bc] A question regarding the proposal for issue # 91

>What is an instance of a type?

I assume he means an object of a type.

>I assumed that P here was a parameter of type 'type',
>say,
>
> parameter type P = real ;
> if (P == real) ...
>
>In that snippet, the two expressions would be == unless the
>value of P were overridden.

But if you had

  real r;
  if (r == real)
  
that wouldn't work well (as Paul pointed out). So you still need $typeof,
so you can say

  real r;
  if ($typeof(r) == real)
  

Steven Sharp
sharp@cadence.com
Received on Wed Nov 24 14:04:18 2004