why copy constructor is not being called?

[suresh]

Hi

Could you please tell why copy constructor is not called in the first
line in main(), as mentioned in the text books. I used g++ version
4.1.2 on debian etch

thanks
suresh

# include <iostream>
using namespace std;
class base
{
char s;
public:
base( ){cout<< "constructi on" << endl;}
base(const char a ) {cout << "constructi on with arg" << endl;}
base(const base & a){cout <<"copy constructor" << endl;}
base & operator=(const base & a){cout << "assignment operator" <<
endl;}
~base( ) {cout << "destructio n" << endl;}
};

main ( )
{
base b1 = 'x'; //why copy constructor is NOT called here as given in
books?
base b2 = b1; //copy constructor is called

}

[ltcmelo]

On Apr 15, 1:15*pm, suresh <suresh.amritap [e-mail] ote:

Hi

Could you please tell why copy constructor is not called in the first
line in main(), as mentioned in the text books. I used g++ version
4.1.2 on debian etch

thanks
suresh

# include <iostream>
using namespace std;
class base
{
* char s;
public:
* base( ){cout<< "constructi on" << endl;}
* base(const char a ) {cout << "constructi on with arg" << endl;}
* base(const base & a){cout <<"copy constructor" << endl;}
* base & operator=(const base & a){cout << "assignment operator" <<
endl;}
* ~base( ) {cout << "destructio n" << endl;}

};

main ( )
{
* base b1 = 'x'; //why copy constructor is NOT called here as given in
books?
* base b2 = b1; //copy constructor is called

}- Hide quoted text -

Hi.

For this particular case your constructor that takes a char as an
argument is actually the one supposed to be called. Why should your
copy constructor be called? You're not creating a copy...
--
Leandro T. C. Melo

[Pete Becker]

On 2008-04-15 12:23:16 -0400, ltcmelo <lt*****@gmail. comsaid:

On Apr 15, 1:15*pm, suresh <suresh.amritap [e-mail] ote:

>Hi

Could you please tell why copy constructor is not called in the first
line in main(), as mentioned in the text books. I used g++ version
4.1.2 on debian etch

thanks
suresh

# include <iostream>
using namespace std;
class base
{
* char s;
public:
* base( ){cout<< "constructi on" << endl;}
* base(const char a ) {cout << "constructi on with arg" << endl;}
* base(const base & a){cout <<"copy constructor" << endl;}
* base & operator=(const base & a){cout << "assignment operator" <<
endl;}
* ~base( ) {cout << "destructio n" << endl;}

};

main ( )
{
* base b1 = 'x'; //why copy constructor is NOT called here as given in

>books?
* base b2 = b1; //copy constructor is called

For this particular case your constructor that takes a char as an
argument is actually the one supposed to be called. Why should your
copy constructor be called? You're not creating a copy...

In fact, it is creating a copy. For

base b1 = 'x';

the compiler should construct a temporary object of type base from the
argument 'x', then copy that temporary into b1. However, the answer to
the original question is that the compiler is permitted to elide the
copy if the copy constructor has no side effects, provided that this
does not turn invalid code into valid code. For example, if the copy
constructor is private (try it!), then this initialization is illegal,
regardless of whether the compiler elides the copy constructor.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

[Rolf Magnus]

Pete Becker wrote:

the compiler should construct a temporary object of type base from the
argument 'x', then copy that temporary into b1. However, the answer to
the original question is that the compiler is permitted to elide the
copy if the copy constructor has no side effects,

Actually, the compiler is also allowed to do this if the constructor does
have side effects. Printing to cout is a side effect.

[ltcmelo]

On Apr 15, 1:32*pm, Pete Becker <p...@versatile coding.comwrote :

On 2008-04-15 12:23:16 -0400, ltcmelo <ltcm...@gmail. comsaid:

On Apr 15, 1:15*pm, suresh <suresh.amritap [e-mail] ote:

Hi

Could you please tell why copy constructor is not called in the first
line in main(), as mentioned in the text books. I used g++ version
4.1.2 on debian etch

thanks
suresh

# include <iostream>
using namespace std;
class base
{
* char s;
public:
* base( ){cout<< "constructi on" << endl;}
* base(const char a ) {cout << "constructi on with arg" << endl;}
* base(const base & a){cout <<"copy constructor" << endl;}
* base & operator=(const base & a){cout << "assignment operator" <<
endl;}
* ~base( ) {cout << "destructio n" << endl;}

};

main ( )
{
* base b1 = 'x'; //why copy constructor is NOT called here as givenin

books?
* base b2 = b1; //copy constructor is called

For this particular case your constructor that takes a char as an
argument is actually the one supposed to be called. Why should your
copy constructor be called? You're not creating a copy...

In fact, it is creating a copy. For

base b1 = 'x';

the compiler should construct a temporary object of type base from the
argument 'x', then copy that temporary into b1. However, the answer to
the original question is that the compiler is permitted to elide the
copy if the copy constructor has no side effects, provided that this
does not turn invalid code into valid code. For example, if the copy
constructor is private (try it!), then this initialization is illegal,
regardless of whether the compiler elides the copy constructor.

Yes, you're correct (thanks for the note). In fact I think I had a
problem with g++ a few years ago because of that. Cant't remember
exactly what it was...
--
Leandro T. C. Melo

[Pete Becker]

On 2008-04-15 12:38:56 -0400, Rolf Magnus <ra******@t-online.desaid:

Pete Becker wrote:

>the compiler should construct a temporary object of type base from the
argument 'x', then copy that temporary into b1. However, the answer to
the original question is that the compiler is permitted to elide the
copy if the copy constructor has no side effects,

Actually, the compiler is also allowed to do this if the constructor does
have side effects. Printing to cout is a side effect.

Yup.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

[Andrey Tarasevich]

suresh wrote:

...
Could you please tell why copy constructor is not called in the first
line in main(), as mentioned in the text books. I used g++ version
4.1.2 on debian etch
...

Because the language specification explicitly allows the compiler to
optimize away the copying of temporary objects, even if the
copy-constructor has any side-effects.

In other words, the constructor might get called. Or it might not get
called. It might depend on the compiler, on the compiler settings and
even on the concrete context in the code. Just don't rely on it.

--
Best regards,
Andrey Tarasevich