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Re: [xsl] Why does the addition of one (1) to a positiveInteger produce an integer?

2010-08-30 02:19:17
Any distinction between types that do not reduce the number of
permissible values is close to a placebo ;-)

One could argue that (guessing!) that in the case of f:iter(ate)
parameters <= 0 could be handled easily by doing nothing. And one
might also claim that Googol or Googolplex are not reasonable inpit
values; hence explicit range checking might be better than relying on
xs:someType,(and result in better error messages, too).

YMMV.
-W

On 29 August 2010 22:46, Michael Kay <mike(_at_)saxonica(_dot_)com> wrote:

When I run this I get an error, "The argument to f:iter is 
xs:positiveInteger, the supplied value is xs:integer"

What is the rationale for this? After all, if I add one (1) to any 
positiveInteger the result must be a positiveInteger.


There might be logic for saying that the result of xs:positiveInteger + 
xs:positiveInteger should always be xs:positiveInteger, but a rule that says 
xs:positiveInteger + xs:integer gives xs:positiveInteger provided that the 
xs:integer is positive (or provided it is greater than the negation of the 
first argument?) would be rather arbitrary.

Generally I would advise against writing a function that expects subtypes of 
xs:integer as an argument, because you will never be able to supply the 
argument as a simple literal. These types were designed for validation, not 
for type-checking programs. That's why, for example, the standard functions 
insert() and remove() declare the expected type as xs:integer rather than 
xs:positiveInteger. (It's also why in XSLT 2.0 types such as 
xs:positiveInteger were excluded from the set of types recognized by a basic 
XSLT processor). If you want to reject negative integers, do it by 
conditional code in the body of the function, not by means of the type 
signature.

Michael Kay
Saxonica

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