On Wed, Feb 1, 2012 at 12:14 PM, Matthieu Monrocq < [email protected]> wrote:
> Le 31 janvier 2012 22:57, Richard Smith <[email protected]> a écrit : > > On Tue, Jan 31, 2012 at 12:40 AM, Abramo Bagnara <[email protected] >> > wrote: >> >>> Il 31/01/2012 03:22, Eli Friedman ha scritto: >>> > On Mon, Jan 30, 2012 at 6:01 PM, Richard Smith <[email protected]> >>> wrote: >>> >> On Mon, Jan 30, 2012 at 4:29 PM, Eli Friedman <[email protected] >>> > >>> >> wrote: >>> >>> >>> >>> On Mon, Jan 30, 2012 at 2:27 PM, Richard Smith >>> >>> <[email protected]> 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
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