Re: relations aren't types?

"John Jacob" <jingleheimerschmitt_at_hotmail.com> wrote in message <news:72f08f6c.0401141943.415dcbb0_at_posting.google.com>...

> The OTHER type IS the scalar type. Dates as structures don't exist,
> only the components of the possible representation, i.e., the year
> component, the month component, etc.,. Dates as strings are not
> dates, they are strings. It's just that an operator exists to extract
> the string representation of a date from a given date value.

We're neither supposed to know nor care how a "scalar" date type is implemented "under the blankets". It could be a structure, a string, a count of days since 1899-12-30, seconds since 1970-01-01 midnight UTC...

> Clearly, we could represent all values with scalar types, but if we
> did that, we would have to write the code for joins and other
> non-scalar operators ourselves. By partitioning values into scalar
> and non-scalar types, we can build a compiler that can write that code
> for us.

It is important to stay out of the Turing tarpit, or you'll wind up doing everything in OISC or URISC, machine languages with one instruction. That's right, one. *Everything* else is syntactic sugar.

--
Joe Foster <mailto:jlfoster%40znet.com>  DC8s in Spaace: <http://www.xenu.net/>
WARNING: I cannot be held responsible for the above        They're   coming  to
because  my cats have  apparently  learned to type.        take me away, ha ha!