Subject: Re: [boost] [config] consexpr workaround
From: Stewart, Robert (Robert.Stewart_at_[hidden])
Date: 2010-11-22 09:09:26
> From: "Stewart, Robert" <Robert.Stewart_at_[hidden]>
> > The purpose of constexpr is to mark something as a compile
> > time constant such that it can be used in other compile time
> > expressions, including types that weren't allowed as compile
> > time constants before. It allows for constant-expression
> > functions, data, and even constructors. Constant-expression
> > data disallows dynamic initialization which const objects can
> > undergo. Without constexpr support, neither
> > constant-expression functions nor constructors are supported,
> > nor is floating point constant-expression data. Thus, const
> > is inappropriate in those cases, so BOOST_CONSTEXPR_OR_CONST
> > cannot be used.
> I think that we need to declare a constant variable as
> constexpr if we want it to be part of other constexpr. Am I wrong?
You're partly right. Compile time constant integral values can be expressed several ways and all can be used in a constexpr. If you need a constexpr for another purpose, then the other approaches don't apply. However, in that case, BOOST_CONSTEXPR_OR_CONST doesn't apply. If that macro does apply, then you don't need constexpr.
> > For non-floating point constant-expression data, const
> > could work fairly well, but it permits dynamic
> > initialization, so for platforms supporting constexpr, code
> > would fail to compile that mistakenly compiled on platforms
> > without constexpr support.
> You are right. But it works well when the the data is
> staticaly initialized, which is the case for most of the constants.
Sure, but you don't need constexpr in that case, so you don't need BOOST_CONSTEXPR_OR_CONST.
> > I don't think BOOST_CONSTEXPR_OR_CONST is helpful.
> Do you mean that a library that must provide a C++03
> interface can not use constexpr on C++0x compilers as the
> meaning can change? In Boost.Ratio "num" and "den" are
> defined as const for C++03 compilers, and need to be
> constexpr for C++0x compilers. How do you write such a code
> in a "portable" way?
They don't need to be constexpr. The C++03 code will work.
Rob Stewart robert.stewart_at_[hidden]
Software Engineer, Core Software using std::disclaimer;
Susquehanna International Group, LLP http://www.sig.com
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