Re: Template conversion operator ambiguity?

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Thu, 22 Jul 2010 01:30:26 -0700 (PDT)
Message-ID:
<75d39062-f66f-4002-8c89-b9ccc1344167@w31g2000yqb.googlegroups.com>
On Jul 22, 12:01 am, Ian Collins <ian-n...@hotmail.com> wrote:

While "porting" some existing code to g++, I ran into
a problem with a generic value class that uses a template
conversion operator to return values. Here's a stripped down
version:

#include <string>

struct Value
{
   Value() {}

   template <typename T> operator T() { return T(); }
};

int main()
{
   Value v;

   int n(v);
   std::string s(v);
}

My original compiler is happy with the string initialisation,
but g++ whinges about the conversion being ambiguous (edited
to cut the noise):

example.cc:15: error: call of overloaded 'basic_string(Value&)' is ambiguous
candidates are:
std::basic_string<_CharT, _Traits, _Alloc>::basic_string(const _CharT*,
const _Alloc&)

std::basic_string<_CharT, _Traits, _Alloc>::basic_string(const
std::basic_string<_CharT, _Traits, _Alloc>&)

std::basic_string<_CharT, _Traits, _Alloc>::basic_string(const _Alloc&)

But it will happily accept

std::string s = v;

So my question is, which compiler is correct?


I'd say g++. In the original code, the compiler is supposed to
convert v to something that can be used to call a constructor of
std::string: since there are several constructors which take
a single argument, the compiler can't choose between them, and
the call is ambiguous.

When you modify the definition to use copy initialization (with
the = sign), the semantics are: convert the expression to
std::string, then use the copy constructor. In this case,
there's only one conversion the std::string which doesn't
involve more than one user defined conversions, so there is no
ambiguity.

--
James Kanze

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