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Re: Various comments



dyb@xxxxxxxxxxxxxx writes:

>> ... So far, in Scheme, I know that (foo ...)
>> is a form that does something, and to find out what it does, I
>> need to know what FOO does. If something does not show up at the
>> first position in a list, it's a variable reference ...
>
> This is true only if you never enclose an expression in a macro call.  If
> you never do, and you never use variable transformers yourself, you won't
> have any worries.

I see your point, thanks for explaining it.

The difference I still see is that the possibility of such a
transformation is signalled by the lexical environment in R5RS,
but not necessarily so in R6RS. Is this desired?

I'm not happy with this, though. My concerns would only be gone if
the system would not allow such transformations to occur outside
of the lexical scope of a macro application. I'll have to think
about whether this is possible to do in a clean way.

Thanks and regards,
        -- Jorgen

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