(j3.2006) derived types with type parameters are different

[Bill Long]

Aleksandar Donev wrote:
> Jim Xia wrote:
>   
>> For two sequence types to be the same type, they 
>> must have the same memory layout for any pair of objects that having the 
>> same type parameter values.
>>     
> Wow, are you serious. Does the IBM compiler now implement a Maple 
> symbolic algebra proof (reduction) system???
>   

Even 40 years ago the IBM compiler supported symbolic algebra (I think 
the extension was called FORMAC).  It would even let you convert the 
resulting expression into a function, compile it, and dynamically relink 
it into your program so you could get numerical values out.

More to the point, most compilers have some level of internal symbolic 
algebra capability for the interpretation and simplification of 
expressions.  At least at the level of knowing that N and N+0 are the 
same.  Not quite as clear for Andy's example of N*(N-1) and N*N-N.

> Seriously, we cannot possibly add ANY burden on compilers due to this 
> "over-integration". The equivalence of SEQUENCE types is an old 
> "feature" which has little to no place in modern programming. Using it 
> with derived type parameters should certainly not be encouraged, yet 
> alone glorified by forcing any compiler to do more work than necessary.
>
> If there is no words in the standard, Robert should submit an interp.
>   
The words I see require that the corresponding components have the same 
attributes. That boils down to whether A(N) and A(N+0) have the same 
dimension attribute.  I suspect the answer is yes, whether that's what 
we intended or not.


> I would say that two separate derived types with parameters are never 
> equivalent even if they are SEQUENCE, even if they are cut-and-pasted.
>   

At least for the cut-and-paste case, that horse / warthog left the barn 
5 years ago.  To late to change it now.

Cheers,
Bill


> Best,
> Aleks
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>   

-- 
Bill Long                                   longb at cray.com
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