destructor invoked without constructor.

[sam_cit]

Hi Everyone,

I have the following unit to explain the problem that i have,

class sample
{
public : sample()
{
printf("in sample...\n");
}
~sample()
{
printf("out sample...\n");
}
void invoke(sample obj)
{
printf("in invoke...\n");
}
};

int main()

{

sample obj;
sample obj1;
obj1.invoke(obj );
printf("...end. ..\n");
return 0;

}
The following is the output,

in sample...
in sample...
in invoke...
out sample...
....end...
out sample...
out sample...

i think the first "out sample..." is for the object obj passed by
value, which indicates that a new temp object has been destroyed,
which means that it should have been created, if so, shouldn't the "in
sample" have been printed once more before "in invoke"...?

[=?iso-8859-1?q?Erik_Wikstr=F6m?=]

On 4 Maj, 11:03, sam_...@yahoo.c o.in wrote:

Hi Everyone,

I have the following unit to explain the problem that i have,

class sample
{
public : sample()
{
printf("in sample...\n");
}
~sample()
{
printf("out sample...\n");
}
void invoke(sample obj)
{
printf("in invoke...\n");
}
};

int main()

{

sample obj;
sample obj1;
obj1.invoke(obj );
printf("...end. ..\n");
return 0;

}

The following is the output,

in sample...
in sample...
in invoke...
out sample...
...end...
out sample...
out sample...

i think the first "out sample..." is for the object obj passed by
value, which indicates that a new temp object has been destroyed,
which means that it should have been created, if so, shouldn't the "in
sample" have been printed once more before "in invoke"...?

It is created, but using the copy-constructor, which you have not
implemented so the compiler-generated one is used, and that one does
not print anything. Add

sample(const sample&) { std::cout << "copy\n"; }

to your class and see what happens.

--
Erik Wikström

[Sylvester Hesp]

<sa*****@yahoo. co.inwrote in message
news:11******** **************@ l77g2000hsb.goo glegroups.com.. .

Hi Everyone,

I have the following unit to explain the problem that i have,

class sample
{
public : sample()
{
printf("in sample...\n");
}
~sample()
{
printf("out sample...\n");
}
void invoke(sample obj)
{
printf("in invoke...\n");
}
};

int main()

{

sample obj;
sample obj1;
obj1.invoke(obj );
printf("...end. ..\n");
return 0;

}
The following is the output,

in sample...
in sample...
in invoke...
out sample...
...end...
out sample...
out sample...

i think the first "out sample..." is for the object obj passed by
value, which indicates that a new temp object has been destroyed,
which means that it should have been created, if so, shouldn't the "in
sample" have been printed once more before "in invoke"...?

No. The temporary is created using the copy constructor, and since you
didn't supply any it was automatically generated (and obviously not with the
printf statement in it). Add the following constructor and it should work as
you expect:

sample(const sample & s)
{
printf("in sample... (copy ctor)\n");
}

Btw, it might be handy to output the this pointer as well so you can exactly
see what ctor-dtor pairs belong to eachother :)

- Sylvester

[WittyGuy]

On May 4, 2:03 pm, sam_...@yahoo.c o.in wrote:

Hi Everyone,

I have the following unit to explain the problem that i have,

class sample
{
public : sample()
{
printf("in sample...\n");
}
~sample()
{
printf("out sample...\n");
}
void invoke(sample obj)
{
printf("in invoke...\n");
}
};

int main()

{

sample obj;
sample obj1;
obj1.invoke(obj );
printf("...end. ..\n");
return 0;

}

The following is the output,

in sample...
in sample...
in invoke...
out sample...
...end...
out sample...
out sample...

i think the first "out sample..." is for the object obj passed by
value, which indicates that a new temp object has been destroyed,
which means that it should have been created, if so, shouldn't the "in
sample" have been printed once more before "in invoke"...?

Invoking a function with pass by value argument would definitely call
copy constructor, whereas pass the argument by reference wouldn't
create an object.
So If you would've defined a copy constructor for the class sample,
within which a print statement your doubt would've got cleared.

I hope this helps.

-
Sukumar R

[sam_cit]

On May 4, 2:21 pm, WittyGuy <wittyguys...@g mail.comwrote:

On May 4, 2:03 pm, sam_...@yahoo.c o.in wrote:

Hi Everyone,

I have the following unit to explain the problem that i have,

class sample
{
public : sample()
{
printf("in sample...\n");
}
~sample()
{
printf("out sample...\n");
}
void invoke(sample obj)
{
printf("in invoke...\n");
}
};

int main()

{

sample obj;
sample obj1;
obj1.invoke(obj );
printf("...end. ..\n");
return 0;

}

The following is the output,

in sample...
in sample...
in invoke...
out sample...
...end...
out sample...
out sample...

i think the first "out sample..." is for the object obj passed by
value, which indicates that a new temp object has been destroyed,
which means that it should have been created, if so, shouldn't the "in
sample" have been printed once more before "in invoke"...?

Invoking a function with pass by value argument would definitely call
copy constructor, whereas pass the argument by reference wouldn't
create an object.
So If you would've defined a copy constructor for the class sample,
within which a print statement your doubt would've got cleared.

I hope this helps.

-
Sukumar R- Hide quoted text -

- Show quoted text -

So it is correct to understand that the default copy constructor would
be invoked in all calls to member functions where the object of the
same class's type is passed by value?
....

Thanks in advance!!!

[sam_cit]

On May 4, 2:14 pm, Erik Wikström <eri...@student .chalmers.sewro te:

On 4 Maj, 11:03, sam_...@yahoo.c o.in wrote:

Hi Everyone,

I have the following unit to explain the problem that i have,

class sample
{
public : sample()
{
printf("in sample...\n");
}
~sample()
{
printf("out sample...\n");
}
void invoke(sample obj)
{
printf("in invoke...\n");
}
};

int main()

{

sample obj;
sample obj1;
obj1.invoke(obj );
printf("...end. ..\n");
return 0;

}

The following is the output,

in sample...
in sample...
in invoke...
out sample...
...end...
out sample...
out sample...

i think the first "out sample..." is for the object obj passed by
value, which indicates that a new temp object has been destroyed,
which means that it should have been created, if so, shouldn't the "in
sample" have been printed once more before "in invoke"...?

It is created, but using the copy-constructor, which you have not
implemented so the compiler-generated one is used, and that one does
not print anything. Add

sample(const sample&) { std::cout << "copy\n"; }

to your class and see what happens.

--
Erik Wikström- Hide quoted text -

- Show quoted text -

And what is the reason that only copy constructor is invoked in this
case and why not the regular constructor?

[Gianni Mariani]

sa*****@yahoo.c o.in wrote:
....

So it is correct to understand that the default copy constructor would
be invoked in all calls to member functions where the object of the
same class's type is passed by value?

Remove the word "default" and then you are correct.

The compiler provides a "default" copy constructor (if one is needed).
The programmer can provide a copy constructor, in which case the
compiler does not create a "default" one. Either way, if you pass by
value, the copy constructor is invoked.

[Gianni Mariani]

sa*****@yahoo.c o.in wrote:
....

And what is the reason that only copy constructor is invoked in this
case and why not the regular constructor?

Because it is defined that way in the standard. Passing by value by
definition requires the compiler to (appear to) create a temporary copy
of the object being passed. New copies of objects are created using the
copy constructor, almost by definition.

[Sylvester Hesp]

<sa*****@yahoo. co.inwrote in message
news:11******** **************@ c35g2000hsg.goo glegroups.com.. .
On May 4, 2:14 pm, Erik Wikström <eri...@student .chalmers.sewro te:

On 4 Maj, 11:03, sam_...@yahoo.c o.in wrote:

It is created, but using the copy-constructor, which you have not
implemented so the compiler-generated one is used, and that one does
not print anything. Add

sample(const sample&) { std::cout << "copy\n"; }

to your class and see what happens.

And what is the reason that only copy constructor is invoked in this
case and why not the regular constructor?

Because you're creating a copy, obviously. That's what the copy constructor
is for. You're not creating a new object using no arguments. How else would
this work:
#include <iostream>

struct sample
{
int i;

sample() : i(0) { }
sample(int _i) : i(_i) { }
};

void foo(sample s)
{
std::cout << s.i << std::endl;
}

int main()
{
sample s(23);
foo(s);
}

If the temporary passed to foo() were to be default-constructed, you are
never able to pass the value 23 to foo. The copy-ctor is invoked to
construct a new sample that is a copy from the sample passed to foo(), so
you want the value 23 to be copied. This is the responsibility from the copy
constructor.

- Sylvester Hesp

[sam_cit]

On May 4, 2:59 pm, "Sylvester Hesp" <s.h...@oisyn.n lwrote:

<sam_...@yahoo. co.inwrote in message

news:11******** **************@ c35g2000hsg.goo glegroups.com.. .
On May 4, 2:14 pm, Erik Wikström <eri...@student .chalmers.sewro te:

On 4 Maj, 11:03, sam_...@yahoo.c o.in wrote:

It is created, but using the copy-constructor, which you have not
implemented so the compiler-generated one is used, and that one does
not print anything. Add

sample(const sample&) { std::cout << "copy\n"; }

to your class and see what happens.

And what is the reason that only copy constructor is invoked in this
case and why not the regular constructor?

Because you're creating a copy, obviously. That's what the copy constructor
is for. You're not creating a new object using no arguments. How else would
this work:

#include <iostream>

struct sample
{
int i;

sample() : i(0) { }
sample(int _i) : i(_i) { }

};

void foo(sample s)
{
std::cout << s.i << std::endl;

}

int main()
{
sample s(23);
foo(s);

}

If the temporary passed to foo() were to be default-constructed, you are
never able to pass the value 23 to foo. The copy-ctor is invoked to
construct a new sample that is a copy from the sample passed to foo(), so
you want the value 23 to be copied. This is the responsibility from the copy
constructor.

- Sylvester Hesp

Thanks you very much. I understand the purpose of copy constructors
being used i would assume they are invoked for the following case too,

foo(s(23));

isn't it?

I also tried to define a constructor like

sample(sample obj) // i get an error saying first parameter
shouldn't be of type sample but the following works fine
sample(int i, sample obj)

why doesn't the first work? I pass an already existing object to the
constructor, a new temp object would be created with the help of copy
constructor and i'm using its value to create my object, what is wrong
in this scenario?

Thanks in advance!!!