initialization of function argument

[subramanian100in]

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

 

[Alf P. Steinbach]

* (e-mail address removed), India:

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 ?

Formally it's copy initializion, per §8.5/12.

If copy initialization happens, then if the copy
ctor is 'explicit', then calling fn(obj) will not
compile. Am I correct ?

Try submitting the following code to Comeau Online,

struct Foo
{
explicit Foo( Foo const& ) {}
};

void bar( Foo ) {}

int main()
{
bar( Foo() );
}

Cheers, & hth.,

- Alf

 

[Mike Wahler]

Suppose we have a class named Test.

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

I'm not sure what you mean by 'direct initialization', but
the above line invokes the copy constructor.

Test copy_init = obj; // copy initialization happens here

This will also invoke the copy constructor.

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?
Yes.

Question is what kind of initialization - direct
initialization or copy initialization, happens to
construct arg ?

Copy construction.

What does the standard say in this regard ?

I don't have my copy at hand.

If copy initialization happens, then if the copy
ctor is 'explicit', then calling fn(obj) will not
compile.

Sure it will.

Am I correct ?

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'

Kindly explain

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

 

[Alf P. Steinbach]

* Mike Wahler:

Copy construction.


I don't have my copy at hand.

See my reply about two hours earlier in this thread. It's often a good
idea to read earlier replies in the thread, before posting.

Sure it will.

I'm sorry, it will only compile with a broken compiler (e.g. MSVC).

No.

He is correct.

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.

When I added the word "explicit" to the copy constructor of your
program, and commented out the copy_init declaration,

Comeau C/C++ 4.3.9 (Mar 27 2007 17:24:47) for ONLINE_EVALUATION_BETA1
Copyright 1988-2007 Comeau Computing. All rights reserved.
MODE:strict errors C++ C++0x_extensions

"ComeauTest.c", line 47: error: class "Test" has no suitable copy
constructor
fn(obj);
^

1 error detected in the compilation of "ComeauTest.c".

g++ also failed to compile, whereas Visual C++ did compile it.

So we can conclude that of the three compilers I tested your code with,
one did not handle the "explicit" copy constructor correctly.


Cheers, & hth.,

- Alf

 

[Erik Wikström]

* Mike Wahler:

See my reply about two hours earlier in this thread. It's often a good
idea to read earlier replies in the thread, before posting.



I'm sorry, it will only compile with a broken compiler (e.g. MSVC).



He is correct.



When I added the word "explicit" to the copy constructor of your
program, and commented out the copy_init declaration,

Comeau C/C++ 4.3.9 (Mar 27 2007 17:24:47) for ONLINE_EVALUATION_BETA1
Copyright 1988-2007 Comeau Computing. All rights reserved.
MODE:strict errors C++ C++0x_extensions

"ComeauTest.c", line 47: error: class "Test" has no suitable copy
constructor
fn(obj);
^

1 error detected in the compilation of "ComeauTest.c".

g++ also failed to compile, whereas Visual C++ did compile it.

You should check which flags you are using then, I could not get that
past the VC++ compiler.

 

[Juha Nieminen]

Test direct_init(obj); // direct initialization happens here
Test copy_init = obj; // copy initialization happens here

Those lines do the exact same thing. The latter is just syntactic
sugar for the former.

 

[Alf P. Steinbach]

* Juha Nieminen:

Those lines do the exact same thing. The latter is just syntactic
sugar for the former.

I'm sorry, that's incorrect.

Try reading the posting you replied to to pick up relevant terms.

Read the rest of the thread for discussion and more information (it's
often a good idea to read the articles in a thread before contributing).


Cheers, & hth.,

- Alf

 

[Juha Nieminen]

struct Foo
{
explicit Foo( Foo const& ) {}
};

void bar( Foo ) {}

int main()
{
bar( Foo() );
}

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?

 

[James Kanze]

Those lines do the exact same thing. The latter is just syntactic
sugar for the former.

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]

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?
Yes.

Question is what kind of initialization - direct
initialization or copy initialization, happens to
construct arg ?

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.

What does the standard say in this regard ?

Exactly what I just said.

If copy initialization happens, then if the copy
ctor is 'explicit', then calling fn(obj) will not
compile. Am I correct ?

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]

You should check which flags you are using then, I could not get that
past the VC++ compiler.

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.

 

[Alf P. Steinbach]

* James Kanze:

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."

This is also the language of the original standard.

It's not a contradiction: "explicit" is a limitation on allowable syntax
for constructor calls, namely that they must be explicit, and is not
concerned with the final semantic effect of a call.

In other words, §8.5 on initialization refers to the semantic effect,
whereas §12.3.1 on explicit constructors refers to which syntax syntax
is allowed.

Only in the cases where explicit constructor call syntax is not
available (the only such case I know is passing an argument by value)
does that limit what can be done, the possible semantics.

> (Of course, it contradicts this for
explicit default constructors in the very next sentence. But
that doesn't affect the copy constructor.)

I'm not sure I understand what you mean here or really what you're
referring to. Possibly you're talking about the exception for default
constructors. If so then yes I agree the wording is not very clear.

But no matter.

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.)

That would introduce another special case (in addition to the default
constructor), and this time on the argument type. Which would either be
inconsistent with effect of -- I'm assuming this is allowed --

template< class T >
explicit MyClass( T const & );

or make this templated non-copy constructor allow different call
/syntax/ dependending on the actual argument /type/.

It's messy. I think perhaps the decision to mix construction and
conversion in one common syntax was not optimal, because it places the
decision of whether you want implicit conversion in one central place,
namely in the class definition. Instead, that decision may conceivably
better be expressed at each place a conversion would put its result,
with no implicit conversion as default. However, I'm not sure. It's
difficult to say without experience with a language that does this.

Cheers, & hth.,

- Alf

 

[subramanian100in]

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

 

[James Kanze]

* James Kanze:

[...]

This is also the language of the original standard.

It's not a contradiction: "explicit" is a limitation on
allowable syntax for constructor calls, namely that they must
be explicit, and is not concerned with the final semantic
effect of a call.

In other words, §8.5 on initialization refers to the semantic effect,
whereas §12.3.1 on explicit constructors refers to which syntax syntax
is allowed.

That does sound like the most reasonable interpretation. (Of
what is written, not of what should be specified.)

Only in the cases where explicit constructor call syntax is
not available (the only such case I know is passing an
argument by value) does that limit what can be done, the
possible semantics.

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?).

[...]

That would introduce another special case (in addition to the default
constructor), and this time on the argument type. Which would either be
inconsistent with effect of -- I'm assuming this is allowed --

template< class T >
explicit MyClass( T const & );

or make this templated non-copy constructor allow different call
/syntax/ dependending on the actual argument /type/.

It's messy.

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 perhaps the decision to mix construction and
conversion in one common syntax was not optimal,

I think that there's more or less general agreement about this
now. It's a historical issue.

 

[Juha Nieminen]

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?

Apparently nobody knows.

 

[Victor Bazarov]

Apparently nobody knows.

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