(j3.2006) is MOD (INT_MIN, -1) legal?

[Van Snyder]

If 7.1.5.2.4p1-2 applies to numeric intrinsic functions as well as to
numeric intrinsic operations, the calculation of MPD(INT_MIN,-1) doesn't
have to overflow.

If those paragraphs don't apply to numeric intrinsic functions, perhaps
they should.

On Mon, 2008-10-27 at 14:12 -0800, Michael Ingrassia wrote:
> Interesting question!
> 
> >MOD(A,P) is interpreted as A - INT ( A / P) * P
> 
> That's not what I read in 13.7.80.  There's no explicit or implicit
> claim that the MOD intrinsic is "interpreted".  Rather, it has a value
> which is mathematically given as A-INT (A/P) * P.  Sounds fussy, but I
> think it matters here.
> 
> >This expression applied on MOD(INT_MIN, -1) ... overflows.
> 
> The reason it matters is that mathematical expressions might not overflow nearly
> so often as computational expressions.
> 
> If "v" is a valuation function which takes Fortran expressions and
> produces mathematical values (say, leftover from some denotational semantics
> paper), then it makes sense that
> 
> 	v(A - INT(A/P)*P) = v(A) - v(INT(A/P)*P)
> 			  = v(A) - v(INT(A/P))*v(P)
> 			  = v(A) - v(A/P)*v(P)       ! 13.7.53 case (i)
> 	
> and for A equal to INT_MIN and P equal to (-1), the Fortran expression
> A/P does overflow on my processor and it might be that mathematically
> v(overflow-expression) is some sort of Infinite value.  At this point
> we need to know the range of the valuation function v.  For integers,
> we haven't blessed any particular model with an Infinity, so I think we're
> stuck concluding that the standard has cunningly failed to specify a
> value for MOD in this case.
> 
> Some ways out:
> 
> 1) we only claim to specify models for inputs that can be described in 
> terms of the Integer model.   On the edges, you're on your own.
> 2) we really didn't mean to make the value depend on whether A/P overflowed.
> we should come up with a better definition.
> 3) we really did mean that MOD can overflow if A/P overflows.  We could add
> a NOTE that says so.
> 
> 	--Michael I.   
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