Re: Ambiguous constructor call

From:
=?ISO-8859-1?Q?Erik_Wikstr=F6m?= <Erik-wikstrom@telia.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 03 Sep 2007 10:25:33 GMT
Message-ID:
<xiRCi.7720$ZA.4003@newsb.telia.net>
On 2007-09-03 11:42, xtrigger303@gmail.com wrote:

On Sep 3, 2:44 am, Markus Schoder <a3vr6dsg-use...@yahoo.de> wrote:

On Sun, 02 Sep 2007 17:58:17 +0000, xtrigger303 wrote:

Hi to all,
I'm working on a smart pointer implementation and I'm trying to get
automatic type conversion between different pointer types. I stumbled
upon something weird (at least for me) I summarized it in the code
below. I was expecting both things at the end to work or not work at
all....
Any insight?
Thanks in advance,
Francesco

#include <iostream>

class A;

//

class B
{
public:

   B() { std::cout << "B()\n"; }

   B( B const & ) { std::cout << "B( B const & )\n"; }

   ~B() { std::cout << "~B()\n"; }

   B & operator=( B const & ) { std::cout << "B & operator=( B const

& )

\n"; return *this; }

   template< typename T >
   operator T() const;
};

//

class A
{
public:

   A() { std::cout << "A()\n"; }

   explicit A( int ) { std::cout << "A( int )\n"; }

   A( A const & ) { std::cout << "A( A const & )\n"; }

   ~A() { std::cout << "~A()\n"; }

   A & operator=( A const & ) { std::cout << "A & operator=( A const

& )

\n"; return *this; }
};

//

template< typename T >
B::operator T() const { std::cout << "B::operator T() const\n"; return
T(); }

//

int main( )
{
   B obj001;
   A obj002 = obj001; // this works


The standard mandates that this behaves as if a temporary A object is
created first from obj001 and then obj002 is copy constructed from this
temporary object (to further complicate things the actual copy
construction may be elided but the compiler must check that it would have
been possible). Creating the temporary object is an implicit conversion
and hence the explicit A(int) is not considered.

   //A obj003( obj001 ); // this is ambiguous


This is an explicit constructor call hence A(int) as well as the copy
constructor are considered.

}


--
Markus Schoder- Hide quoted text -

- Show quoted text -


Thanks, I think I got it.
But what if I remove the explicit on the constructor.
Does the compiler choose the "shortest" implicit conversion?


No, when writing 'A obj002 = obj001;' it means that the copy constructor
will be used (even though it might be elided), so the other constructors
do not matter.

--
Erik Wikstr?m

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