From: Larry Evans (cppljevans_at_[hidden])
Date: 2008-03-24 17:22:30
On 03/24/08 15:16, Eric Niebler wrote:
> Larry Evans wrote:
>> sort of succinct formula. My last attempt was:
> I confess I don't understand your notation in that message.
The notation is supposed to be a language grammar similar to that
shown in compiler texts, except it doesn't surround the grammar
terminals with quotes. So, to make it closer to that, use:
= 'terminal::type' //rule0
| (expr_type+)'::type' //rule1
and where the postfix + means 1 or more of the preceding. So, one
possible instance of sentence satisfying this grammar is (tracing the
derivations and subscripting different instances of grammar symbols with
_i, where i is an positive integer, and showing the derivations steps
with -(ruleX->, where ruleX is one of the comments labelling
the grammar altenatives in the above equattion):
-rule1-> (expr_type_1, expr_type_2, expr_type_3)'::type'
-rule0-> ('terminal::type'_1 'terminal::type'_2
Now, the terminal symbol, 'terminal::type' is supposed to represent
any proto::terminal<> grammar and the appending '::type' is supposed
to represent the conversion of the proto::grammar to the corresponding
expression. The (expr_type+) is supposed to represent any other
proto "expression" grammar. The reason expr<Tag,expr_type+,Size> was
not used was because that's just the concrete syntax version of
expr_type+; hence, just detail. A concrete instance of this would be:
where the concrete symbols corresponding to the last step in the example
abstract derivation above are:
== //none because shift_right is binary
> I also don't
> understand what you're trying to get at with morphisms, a concept I
> don't grasp. Do you think you could try again to explain what you're
> trying to achieve?
Before I try that, I better make sure the above is clear enough.
Could you let me know if it's not clear.
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