initialization of function argument

Suppose we have a class named Test.

Test obj; // assuming default ctor is available
Test direct_init(obj); // direct initialization happens here
Test copy_init = obj; // copy initialization happens here

Suppose we have a function

void fn(Test arg);

When we call fn() with fn(obj), the copy ctor
is called to construct 'arg'. Am I correct?
Question is what kind of initialization - direct
initialization or copy initialization, happens to
construct arg ?
What does the standard say in this regard ?
If copy initialization happens, then if the copy
ctor is 'explicit', then calling fn(obj) will not
compile. Am I correct ?

Kindly explain

Thanks
V.Subramanian

re: initialization of function argument

<subramanian100in@yahoo.comwrote in message
news:84838f64-2f2a-4d75-bb7e-a3d7af660e4d@e23g2000prf.googlegroups.com...I'm not sure what you mean by 'direct initialization', but
the above line invokes the copy constructor.
This will also invoke the copy constructor.
Yes.
Copy construction.
I don't have my copy at hand.
Sure it will.
No.

However, if the copy constructor is declared explicit,
the line

Test copy_init = obj;

will not compile. (VC++ 2005 says:
class 'Test' : no copy constructor available or
copy constructor is declared 'explicit'

Run my example below.


#include <iostream>

class Test
{
static int i;

public:
Test()
{
++i;
std::cout << "Obj " << i << " Default constructor\n";
}

Test(const Test& t)
{
++i;
std::cout << "Obj " << i << " Copy constructor\n";
}

Test& operator=(const Test& t)
{
++i;
std::cout << std::cout << "Obj " << i << " Assignment operator\n";
return *this;
}

~Test()

{
std::cout << "Obj " << i << " destructor\n";
--i;
}
};

int Test::i;

void fn(Test arg)
{
}

int main()
{
Test obj;
Test direct_init(obj);
Test copy_init = obj;
std::cout << "Before calling fn\n";
fn(obj);
std::cout << "After fn()\n";
return 0;
}


Output:

Obj 1 Default constructor
Obj 2 Copy constructor
Obj 3 Copy constructor
Before calling fn
Obj 4 Copy constructor
Obj 4 destructor
After fn()
Obj 3 destructor
Obj 2 destructor
Obj 1 destructor

-Mike

re: initialization of function argument

On 2007-11-25 17:21, Alf P. Steinbach wrote:You should check which flags you are using then, I could not get that
past the VC++ compiler.

--
Erik Wikström

re: initialization of function argument

subramanian100in@yahoo.com, India wrote:Those lines do the exact same thing. The latter is just syntactic
sugar for the former.

re: initialization of function argument

Alf P. Steinbach wrote:Does this mean that making the copy constructor of the class explicit
is an effective way of prohibiting objects of that type from being
passed to functions by value?

If this is so, then calling bar() is basically impossible?

re: initialization of function argument

On Nov 25, 7:18 pm, Juha Nieminen <nos...@thanks.invalidwrote:Not if Test is a class type and obj has a type other than Test.
In that case, the first says to construct a Test with obj, the
second to convert obj to a Test, and construct a Test with the
results of the conversion. In the first, there is no
conversion, so constructors marked explicit can be used, but
conversion operators in the source type can't be), and no copy,
so it is legal even if Test doesn't have an accessible copy
constructor. In the second, there is conversion, so
constructors marked explicit cannot be used, but conversion
operators in the source type can be, and copy, so an accessible
copy constructor must be available.

If obj is of type Test, of course, or a class derived from Test
(ignoring any const or volatile qualifiers), then copy
initialization is defined to have the same behavior as direct
initialization. (It's not really clear if the copy constructor
is declared explicit, of course---the standard contradicts
itself in that case, saying elsewhere that the direct
initialization syntax cannot be used if the constructor is
declared explicit.)

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
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re: initialization of function argument

On Nov 25, 3:14 pm, "subramanian10...@yahoo.com, India"
<subramanian10...@yahoo.comwrote:Yes.
Formally, it's copy initialization. But the standard also says
that if the type of the initialization is the same as, or a
class derived from, the target type, copy initialization behaves
the same as direct initialization. There is only a difference
when conversions are involved.
Exactly what I just said.
I don't think so, but the original standard is far from clear.
In the section on initialization, it makes it very clear that if
the initialization expression in copy initialization has the
same type as, or a type derived from, the target type, the
semantics are the same as those of direct initialization. In
the section concerning explicit, on the other hand, the latest
draft says that an explicit constructor will only be used "where
the direct-initialization syntax (8.5) or where casts (5.2.9,
5.4) are explicitly used." (Of course, it contradicts this for
explicit default constructors in the very next sentence. But
that doesn't affect the copy constructor.)

The simplest solution is just to never declare the copy
constructor explicit. (Since explicit is designed to prevent
conversions, and the copy constructor doesn't "convert" anything
in the classical sense of the word, the logical decision would
have been for explicit to be ignored on the copy constructor.)

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
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re: initialization of function argument

On Nov 25, 6:04 pm, Erik Wikström <Erik-wikst...@telia.comwrote:Also, perhaps, the version. I think that there was some
discussion concerning what was really meant in the standards
committee; the standard does contradict itself on this subject,
and it's quite possible that compiler implementors interpreted
it differently at one time.

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
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re: initialization of function argument

On Nov 26, 3:52 pm, "Alf P. Steinbach" <al...@start.nowrote:

Hello Alf P. Steinbach,

I got clarified with your answer in comp.lang.c++ for the same
question. I thank you for that.
But I had sent this question to comp.std.c++ three days
ago. Since I didn't get the question posted in
comp.std.c++, I thought it got rejected by the moderator
in comp.std.c++. So I posted the same question in
comp.lang.c++.

I regret if I caused any inconvenience to you.

Hope this clarifies.

Thanks
V.Subramanian

re: initialization of function argument

On Nov 26, 11:52 am, "Alf P. Steinbach" <al...@start.nowrote:[...]That does sound like the most reasonable interpretation. (Of
what is written, not of what should be specified.)
There is no such thing as an "explicit constructor call syntax".
Or are you using this as a synonym for "direct-initialization
syntax or casts"?

At any rate, another case would be return values, and I think
some cases involved in exceptions (catching by value, certainly,
but what about the copy which takes place during a throw?).

[...]Yup. More than I realized at first glance. On the other hand,
it would allow the simple rule of always declaring constructors
explicit, without worrying about whether they might be used for
copy or not.

Your template example suggests an even more pernitious example
to me:

class Toto
{
public:
Toto( Toto& other ) ;
template< typename T >
explicit Toto( T const& other ) ;
} ;

void f( Toto ) ;

void
g()
{
Toto t1 ;
Toto const t2 ;
f( t1 ) ; // no problem...
f( t2 ) ; // illegal! even though a copy is passed.
}

Off hand, I'd say that your example with the template is even
more of an argument that something isn't right. That explicit
should probably only have an effect when an implicit conversion
is involved, and should be ignored otherwise. (But I'm not sure
how to express that in standardese.)
I think that there's more or less general agreement about this
now. It's a historical issue.

--
James Kanze (GABI Software) email:james.kanze@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
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re: initialization of function argument

Juha Nieminen wrote:Apparently nobody knows.

re: initialization of function argument

Juha Nieminen wrote:I know.

Yes, it is impossible. The absence of the default c-tor in 'Foo'
makes it impossible to create an instance of 'Foo' to begin the
copying process. And it has nothing to do with the fact that the
copy-c-tor is declared 'explicit'.

V
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