Details

Type: Improvement

Status: Closed

Priority: Minor

Resolution: Fixed

Affects Version/s: V4.0_OS

Fix Version/s: V4.0_ERRATA03

Component/s: URL Conventions

Labels:None

Environment:
Applied

Proposal:

Resolution:
Description
URL Conventions spec section 5.1.1.10 "Numeric Promotion" states:
"If the result of an arithmetic operation does not fit into the type determined by the above rules, the nextwider type is used in the above order, with Edm.Double considered widest."
We may group OData numeric types into three families:
integer types: Byte, SByte, Int16, Int32, Int64
decimal types: Decimal
floating types: Single, Double
Numeric promotion as a result of overflow is undesirable when it results in promotion between number type families. The semantics of arithmetic operators can therefore not be determined by static analysis. For example:
First issue:

If A, B and C integer values, then "(A add B) div C" is an integer unless "A add B" overflows Int64, in which case "(A add B) div C" will be decimal.
Now consider "((A add B) div C) mod D". If "(A add B) div C" is decimal due to overflow, then "((A add B) div C) mod D" is an invalid expression since the "mod" operator is only defined for integral types.
Second issue:

Now consider that X and Y are decimal values. If (X add Y) overflows the range for decimal values, then according to the spec, (X add Y) is promoted to type Single. Two issues then arise:
(1) Type Single has much less precision than either Decimal or Double. The implicit promotion on overflow should (if we permit promotion across number type familes) arguably be to type Double, to avoid unnecessary loss of precision.
(2) "(X add Y)" div 0 is an error if "X add Y" does not overflow, but is positive or negative infinity if "X add Y" overflows and is promoted to a floating type. So the semantics of division are not statically determinable.
Third issue:

Consider (E add F) where E and F are type Single.
Promotion of the result from Single to Double (on overflow, say if the result exponent is outside the range of type Single) is not in accordance with IEEE floating point rules, which specify positive or negative infinity in this case.