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

Matthieu Monrocq matthieu.monrocq at gmail.com
Thu Feb 2 10:40:01 PST 2012


Le 1 février 2012 23:34, Richard Smith <richard at metafoo.co.uk> a écrit :

> 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
>

Not at all :) I hope Clang keeps aiming at maximum standard conformance.

I was reacting to this (from you):

> 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.

I disagreed here on what is the range of representable values of an
enumeration type and gave the formula to compute it: it is slightly broader
than [Min...Max].

> 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.

Which here will affect which values are out-of-range and thus non-constant,
I think.

Of course, I am no expert, so you might want to double check.

The unspecified behavior I was referencing, and which may have confused
you, was more related to runtime. With hindsight I guess I had better not
written about it as it muddled my message.

-- Matthieu
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