🚀 go-pugleaf

In article 2E03@cs.tu-berlin.de, Roman Lechtchinsky <wolfro@cs.tu-berlin.de> writes:
>Steve Clamage wrote:
>>
>> Suppose a copy constructor T::T(const T&) is viewed as a conversion
>> from T to T. If such a conversion is "needed", the shortest conversion
>> sequence must be chosen. That would be the null sequence, and so the
>> copy ctor would never be used or even considered for that purpose.
>> (Just as the sequence int->Rational->float would never be considered in
>> converting an int to a float.)
>
>I'm not too sure here. First, if the copy constructor is the identity 
>conversion, why is it called at all? I mean, if the identity conversion is 
>eliminated from the conversion sequence, why isn't the call to the copy 
>constructor itself eliminated? Second, if the copy constructor is called, why 
>is it called only once? Consider
>
>class A { A(const A&); };
>void foo(A);
>A a;
>
>Now, foo(a) is implicitly converted to foo(A(a)). Why not to foo(A(A(a))? 

Let's not confuse type conversion with making a copy. In the call foo(a)
no type conversion is required. Function foo requires an A object, and is
invoked with an A object. But foo does require a copy of the actual argument,
no matter what its origin. In this case foo gets a local copy of 'a'.

Example 2:
	class B { public: operator A(); };
	B b;
	foo(b);
We do need a conversion of 'b' to type A, which is available. A temporary A
object is created by "B::operator A()", and a copy of that is passed to foo.

In both examples a copy of an A object is needed, accomplished as always
by the copy constructor.

---
Steve Clamage, stephen.clamage@eng.sun.com
---
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