Bug#538647: closed by Florian Weimer <fw@deneb.enyo.de> (Re: Bug#538647: g++-4.3: g++ examines second value to "?" always if it is a template)

[Florian Weimer <fw@deneb.enyo.de>]

* Philip Ashmore:

>> * Philip Ashmore:
>>
>>   
>>> /*HERE*/enum { value = (wanted <= guess) ? result : next_value_type::value };
>>>     
>>
>> Not a bug.  You need to implement your own conditional operator at the
>> template level to make this work.  The version with ?: is not valid
>> C++

> Not valid C++ ? My C++ compiler (g++) disagrees.

My wording was a bit careless.  Technically, it is a valid C++ program
which cannot be compiled by a conforming implementation, no matter
what its implementation limits regarding nested template
instantiations are.  However, for practical purposes, it's still
invalid code.

> I don't remember reading anything about "conditional operator at the
> template level" in Stroustrup's "C++ programming language (third
> edition).

It's something like this:

  template <bool Test, typename True, typename False> struct ite { };
  template <typename True, typename False> struct ite<true, True, False>
    { typedef True value; };
  template <typename True, typename False> struct ite<false, True, False>
    { typedef False value; };

I'm no longer sure that it would help the way you structured your
example.

Here's something instead that works:

  template <unsigned N> struct log2 {
    enum { value = log2<(N >> 1)>::value + 1 };
  };
  
  template <> struct log2<0> {
    enum { value = -1 };
  };

The difference is that template matching stops the recursion.  Plain
expressions don't, even if short-cut operators such as &&, ||, and ?:
are involved---at least in ISO C++ (previous language versions and
their implementations were different).  This is one reason why
template metaprogramming is so weired.