Re: Where in the XPath 2.0 is the text stating that a

> I don't think there's anywhere in the spec that says "in natural language"
> that with the expression E1 < E2, the operand E2 can be a function call;
and
> I don't think there should be.

I differ from that opinion: the ability to specify a function call as
a location step is one of the radical changes from XPath 1.0 to XPath
2.0 and right now it is hidden in the grammar and requires a
three-step deduction from the grammar rules.

The result -- almost three years after the XPath 2.0 spec was
published  -- is that very few people realize they can specify a
function call where a location step is expected.

While the specification should not be a tutorial, groundbreaking
changes and the most powerful new features should be at least
summarized in a special section of the document so that this could be
referenced when answering a question. When answering a question it
isn't at all convincing to ask the user to follow a traversal of
grammar rules in order to show that a major new feature really exists.
Even a single example in the spec. would also be of great help and a
step forward.


--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.
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To invent, you need a good imagination and a pile of junk
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Never fight an inanimate object
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You've achieved success in your field when you don't know whether what
you're doing is work or play



On Thu, Jul 29, 2010 at 2:00 AM, Michael Kay <mike@xxxxxxxxxxxx> wrote:
> XPath 2.0 changed the syntax so that "/" behaved like a proper orthogonal
> binary operator by allowing any expression (subject only to precedence and
> type rules) as its operands. It is of course a higher-order operator in
that
> it evaluates the right-hand operand repeatedly.
>
> Quite properly, the semantics of E1/E2 are explained in terms that are
> independent of the syntactic form of E1 and E2, and it's the grammar rules
> that tell you what kind of expressions are allowed in E1 and E2. This is in
> my view the correct way to describe an orthogonal functional language.
>
> I don't think there's anywhere in the spec that says "in natural language"
> that with the expression E1 < E2, the operand E2 can be a function call;
and
> I don't think there should be. People should expect to be able to use
> function calls anywhere an expression is allowed, and the spec should only
> draw notice to something that's an exception to the general rule.
>
> When writing a tutorial, of course, different rules apply. In my book,
> there's extensive discussion of the binary "/" operator on pages 602-604.
> But the spec is not a tutorial.
>
> Michael Kay
> Saxonica
>
>
> On 29/07/2010 06:11, Dimitre Novatchev wrote:
>>
>> I can see from the productions of the XPAth 2.0 grammar that:
>>
>> B  A StepExpr can be a FilterExpr B  B  B  (27)
>>
>> and
>>
>> B A FilterExpr can be PrimaryExpr B  B  B (38)
>>
>> and
>>
>> B A PrimaryExpr can be FunctionCall B (41)
>>
>>
>> However I cant see anywhere a description in natural language that a
>> step can be a function call, and I think it is important that this
>> fact be stated clearly enough.
>>
>> Am I missing this text?