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

[Michael Kay <mike@xxxxxxxxxxxx>]

> 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