[cfe-commits] r149286 - in /cfe/trunk: lib/AST/ExprConstant.cpp lib/Sema/SemaOverload.cpp test/CXX/expr/expr.const/p2-0x.cpp

[Richard Smith]

On Wed, Feb 1, 2012 at 12:14 PM, Matthieu Monrocq <
matthieu.monrocq at gmail.com> wrote:

> Le 31 janvier 2012 22:57, Richard Smith <richard at metafoo.co.uk> a écrit :
>
> On Tue, Jan 31, 2012 at 12:40 AM, Abramo Bagnara <abramo.bagnara at gmail.com
>> > wrote:
>>
>>> Il 31/01/2012 03:22, Eli Friedman ha scritto:
>>> > On Mon, Jan 30, 2012 at 6:01 PM, Richard Smith <richard at metafoo.co.uk>
>>> wrote:
>>> >> On Mon, Jan 30, 2012 at 4:29 PM, Eli Friedman <eli.friedman at gmail.com
>>> >
>>> >> wrote:
>>> >>>
>>> >>> On Mon, Jan 30, 2012 at 2:27 PM, Richard Smith
>>> >>> <richard-llvm at metafoo.co.uk> wrote:
>>> >>>> Author: rsmith
>>> >>>> Date: Mon Jan 30 16:27:01 2012
>>> >>>> New Revision: 149286
>>> >>>>
>>> >>>> URL: http://llvm.org/viewvc/llvm-project?rev=149286&view=rev
>>> >>>> Log:
>>> >>>> constexpr: disallow signed integer overflow in integral conversions
>>> in
>>> >>>> constant
>>> >>>> expressions in C++11.
>>> >>>
>>> >>> Standard citation?  As far as I can tell, the result of
>>> >>> (int)0x80000000u is implementation-defined, but it's still a constant
>>> >>> expression given how we define it.
>>> >>
>>> >>
>>> >> Oops, r149327. This was (incorrectly) factored out of another change
>>> which
>>> >> I'm still questioning... Consider:
>>> >>
>>> >>   enum E { n = 2 };
>>> >>   E e = (E)5;
>>> >>
>>> >> 5 is not in the range of values of the enumeration (which is 0..3 by
>>> >> [dcl.enum]p7), but is clearly in the underlying type. Is this value
>>> in the
>>> >> range of representable values for its type (or is this undefined
>>> behavior by
>>> >> [expr]p4)?
>>> >
>>> > I think the relevant passage is actually [expr.static.cast]p10:
>>> >
>>> > A value of integral or enumeration type can be explicitly converted to
>>> > an enumeration type. The value is unchanged if the original value is
>>> > within the range of the enumeration values (7.2). Otherwise, the
>>> > resulting value is unspecified (and might not be in that range).
>>> >
>>> >
>>> > That doesn't sound like undefined behavior to me.
>>>
>>> Yes, you're definitely right from a standard point of view, but using
>>> the point of view of constant evaluator, does it make a difference?
>>>
>>> I.e., the conversion of an integer out of enum range specified by
>>> [decl.enum]p7 to that enum type should be a known constant?
>>>
>>> I don't think so, but I'd like to hear your opinion.
>>
>>
>> In C++11, all conditional-expressions are core constant expressions,
>> except those explicitly blacklisted in [expr.const]p2. The only relevant
>> exemption there is "a result that is not mathematically defined or not in
>> the range of representable values for its type;" (which DR1313 generalizes
>> to "an operation that would have undefined behavior").
>>
>> The standard does not make obvious what it means by "the range of
>> representable values" for an enumeration type. Is it the range of values of
>> the enumeration, or is it the range of representable values of the
>> underlying type, or something else? And, when casting an out-of-range value
>> to an enumeration, can the resulting unspecified value be out of the range
>> of representable values for the type?
>>
>> To address the first question, [class.bit]p4 says:
>>
>> "If the value of an enumerator is stored into a bit-field of the same
>> enumeration type and the number of bits in the bit-field is large enough to
>> hold all the values of that enumeration type (7.2), the original enumerator
>> value and the value of the bit-field shall compare equal."
>>
>> From this, and the behavior of integral promotions on enumerations, I
>> believe we can conclude that the range of representable values of an
>> enumeration type is the range of values of the enumeration.
>>
>> Eli's quotation states that casting an out-of-range value to an
>> enumeration produces a value which need not be in the range of values of
>> the enumeration. Therefore, if the unspecified value is not in that range
>> (which by [expr.static.cast]p10 it might not be), then behavior is
>> undefined, and the result of the cast is not a constant expression.
>>
>> That said, such deductive reasoning applied to the (sadly, often
>> imprecise and inconsistent) standard wording has led me to unintended
>> conclusions several times before, so I'm still not certain whether such
>> cases should be constant expressions.
>>
>> - Richard
>>
>>
>> As far as the interpretation of "If the value of an enumerator is stored
> into a bit-field of the same enumeration type and the number of bits in the
> bit-field is large enough to hold all the values of that enumeration type
> (7.2), the original enumerator value and the value of the bit-field shall
> compare equal." I always understood:
>
> Given Min the smallest enumerator value and Max the biggest enumerator
> value of the enumeration:
>
> - if Min is negative and Max positive, then we deduce M = max(abs(Min),
> abs(Max)) and the bitfield should be *just* enough to represent all values
> in [0, M] + a bit sign.
> - if both Min and Max are negative, then the bitfield should be *just*
> enough to represent all values in [0, -Min] + a bit sign.
> - if both Min and Max are positive, then the bitfield should be *just*
> enough to represent all values in [0, Max] (no bit sign).
>
> So, you are looking for K such that 2^(K-1) <= max(abs(Min), abs(Max)) <
> 2^K and all values in [0, 2^K-1] are representable, as well as those in
> [-2^K, 0] if any enumerator was negative.
>
>
> Summary:
> * Min = min { enumerators }
> * Max = max { enumerators }
> * K positive such that 2^(K-1) <= max(abs(Min), abs(Max)) < 2^K
>
> If Min < 0, then the representable range is [-2^K, 2^K-1], else it is [0,
> 2^K-1].
>
> Any value outside this range cannot reliably represented, and it is
> unspecified what happens if one tries to assign such a value. The
> underlying type does not come into play, in the standard.
>
>
> Now, as a compiler implementation, I think that Clang can make the
> additional guarantee (*) that as long as it fits into the underlying type,
> it's okay (it's probably what gcc does, so consistency is good) and that
> otherwise it just "wraps around" (not sure if it should for a signed
> underlying type...), this is the freedom that "unspecified behavior" gives
> us.
>
> And if my reasoning is not totally skewed, then it means this *can* be
> treated as a constant expression.
>

If I understand you correctly, you seem to be assuming that we can just
decide for ourselves what is and is not a constant expression? We cannot;
this is specified (though, in this case, not clearly) by the language
semantics, and a non-conforming implementation (whether it treats too many,
or too few, expressions as constant expressions) will reject valid code,
accept invalid code, and give the wrong semantics to some code.

- Richard
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.llvm.org/pipermail/cfe-commits/attachments/20120201/d72ff297/attachment.html>