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[Bug 6212] [FO] Semantics of idiv

From: <bugzilla@wiggum.w3.org>
Date: Tue, 25 Nov 2008 18:35:49 +0000
To: public-qt-comments@w3.org
Message-Id: <E1L52lJ-0000kt-Oy@farnsworth.w3.org>

http://www.w3.org/Bugs/Public/show_bug.cgi?id=6212





--- Comment #2 from Michael Kay <mike@saxonica.com>  2008-11-25 18:35:49 ---
Looking at the current text:

<quote>
Summary: This function backs up the "idiv" operator and performs an integer
division: that is, it divides the first argument by the second, and returns the
integer obtained by truncating the fractional part of the result. The division
is performed so that the sign of the fractional part is the same as the sign of
the dividend.

If the dividend, $arg1, is not evenly divided by the divisor, $arg2, then the
quotient is the xs:integer value obtained, ignoring (truncating) any remainder
that results from the division (that is, no rounding is performed). Thus, the
semantics " $a idiv $b " are equivalent to " ($a div $b) cast as xs:integer "
except for error situations.
</quote>

we seem to have a long sequence of separate attempts to specify the function,
each trying to say the same thing in different words, but none of them
achieving real precision.

I think the following text does the job more cleanly:

<quote>
Summary: This function backs up the "idiv" operator by performing an integer
division.

If the divisor is (positive or negative) zero, then an error is raised
[err:FOAR0001]. If either operand is NaN or if $arg1 is INF or -INF then an
error is raised [err:FOAR0002].

Otherwise, subject to limits of precision and overflow/underflow conditions,
the result is the largest (furthest from zero) xs:integer value $I such that
fn:abs($I * $arg2) <= fn:abs($arg1) and fn:compare($I * $arg2, 0) =
fn:compare($arg1, 0).

The implementation may adopt a different algorithm provided that it is
equivalent to this formulation in all cases where implementation-dependent or
implementation-defined behaviour does not affect the outcome, for example, the
implementation-defined precision of the result of xs:decimal division.

Note: 

Except in situations involving errors, loss of precision, or
overflow/underflow, the result of ($a idiv $b) is the same as (($a div $b) cast
as xs:integer).

The semantics of this function are different from integer division as defined
in programming languages such as Java and C++.
</quote>


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Received on Tuesday, 25 November 2008 18:39:24 UTC

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