Why this behaviour for literal types?

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From	Andrzej Krzemieński<akrzemi1@gmail.com>
Newsgroups	comp.std.c++
Subject	Why this behaviour for literal types?
Date	Wed, 14 Mar 2012 13:33:54 -0700 (PDT)
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Hi,
This is my literal type:

   struct C
   {
     double m;
     constexpr C( double m ) : m{m} {};
     constexpr double get() const { return m; };
   };

Therefore the following code works:

   constexpr C c = C{5.};
   constexpr double k = c.get();

But the following does not:

   const C cc = C{5.};  // but not constexpr
   constexpr double kk = cc.get();

Why is this so? Object cc is subject to const initialization anyway. It is possible to make it work only by a seemingly irrelevant change in declaration (add constexpr). What was the reason for disallowing this "implicit generation of compile-time constants"?

Regards,
&rzej


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Why this behaviour for literal types? Andrzej Krzemieński<akrzemi1@gmail.com> - 2012-03-14 13:33 -0700
  Re: Why this behaviour for literal types? Daniel Krügler<daniel.kruegler@googlemail.com> - 2012-03-15 11:33 -0700
    Re: Why this behaviour for literal types? Andrzej Krzemieński <akrzemi1@gmail.com> - 2012-03-17 00:53 -0700
      Re: Why this behaviour for literal types? Jason McKesson<jmckesson@gmail.com> - 2012-03-17 07:50 -0700
      Re: Why this behaviour for literal types? Daniel Krügler<daniel.kruegler@googlemail.com> - 2012-03-18 11:31 -0700

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