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[QUOTE="James Kanze, post: 3524547"]
No.  Any user defined constructor, including a copy constructor,
inhibits automatic generation of the default constructor.

A copy constructor is generated any time there is no user
declared copy construtor.  A copy assignment operator is
generated any time there is no user declared copy assignment
operator.  A copy constructor for a class X is a non-template
constructor having an X&, X const&, X volatile& or X const
volatile& as its first parameter, and having default arguments
for all other parameters, if any.  A copy assignment operator is
a non-template assignment operator having an X, X&, X const&, X
volatile& or X const volatile& as parameter.  Note that:

-- a member function template template is never considered a
    copy constructor or copy assignment operator (and thus will
    never prevent the compiler from generating one);

-- you can declare more than one copy constructor or copy
    assignment operator; and

-- operator overload resolution is *always* used when deciding
    what constructor to call, even when "copying", and operator
    overload resolution can choose a constructor which is not a
    copy constructor to copy.

Consider:

class C
    {
    public:
        template< typename T >
        C( T& other ) ;
    } ;

The compiler will generate a C::C( C const& ) copy constructor,
which will participate in operator overload resolution:

C c1 ;
    C c2( c1 ) ;    //  calls the template constructor.
    C c3( C() ) ;   //  calls the compiler generated copy
                    //  constructor.
[/QUOTE]

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