automatic type conversion

[Rahul]

Hi Everyone,

I have the following code,


class B;

class A
{
public : operator B();
};

class B
{
public : B()
{
}
B(const A& ref)
{
printf("destina tion conversion\n");
}

B(const B& ref)
{
printf("copy constructor\n") ;
}
};

A::operator B()
{
printf("source conversion\n");
return B();
}
int main()
{
A obj;
B obj1 = obj;
return(0);
}

so the output is

destination conversion.

There are actually two ways to convert from obj to obj1
1) destination conversion, constructor function of B
2) source conversion, operator function of A and then copy constructor
of B

The compiler has selected the first option, where as i expected a
ambiguity compile time error, what does the standard say for this?

[Victor Bazarov]

Re: automatic type conversion

Rahul wrote:

Hi Everyone,
>
I have the following code,
>
>
class B;
>
class A
{
public : operator B();
};
>
class B
{
public : B()
{
}
B(const A& ref)
{
printf("destina tion conversion\n");
}
>
B(const B& ref)
{
printf("copy constructor\n") ;
}
};
>
A::operator B()
{
printf("source conversion\n");
return B();
}
int main()
{
A obj;
B obj1 = obj;
return(0);
}
>
so the output is
>
destination conversion.
>
There are actually two ways to convert from obj to obj1
1) destination conversion, constructor function of B
2) source conversion, operator function of A and then copy constructor
of B
>
The compiler has selected the first option, where as i expected a
ambiguity compile time error, what does the standard say for this?

Your declaration/definition/initialisation statement

B obj1 = obj;

is actually another form of writing

B obj1((B(obj)));

(and the copying from the temporary of type B to 'obj1' is allowed
to be optimized away by the compiler). As you can see, the temporary
is created from 'obj'. There are two ways to create it: by means of
the converting constructor in B or by means of creating *another*
temporary from A::operator B() and then copy-constructing the original
temporary. The compiler chooses the short way, I guess.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Rahul]

Re: automatic type conversion

On Dec 11, 11:33 pm, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:

Rahul wrote:

Hi Everyone,

>

I have the following code,

>

class B;

>

class A
{
public : operator B();
};

>

class B
{
public : B()
{
}
B(const A& ref)
{
printf("destina tion conversion\n");
}

>

B(const B& ref)
{
printf("copy constructor\n") ;
}
};

>

A::operator B()
{
printf("source conversion\n");
return B();
}
int main()
{
A obj;
B obj1 = obj;
return(0);
}

>

so the output is

>

destination conversion.

>

There are actually two ways to convert from obj to obj1
1) destination conversion, constructor function of B
2) source conversion, operator function of A and then copy constructor
of B

>

The compiler has selected the first option, where as i expected a
ambiguity compile time error, what does the standard say for this?

>
Your declaration/definition/initialisation statement
>
B obj1 = obj;
>
is actually another form of writing
>
B obj1((B(obj)));
>
(and the copying from the temporary of type B to 'obj1' is allowed
to be optimized away by the compiler). As you can see, the temporary
is created from 'obj'. There are two ways to create it: by means of
the converting constructor in B or by means of creating *another*
temporary from A::operator B() and then copy-constructing the original
temporary. The compiler chooses the short way, I guess.
>
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

But the moment i have the following code,

A obj;
B obj1;
obj1 = obj;

I get an error, and the compiler is not able to decide between the
constructor conversion and the operator conversion... So i expected a
similar error in the initial posted code... Does anyone know what the
standard says for these cases?

[red floyd]

Re: automatic type conversion

Rahul wrote:

>
class B;
>
class A
{
public : operator B();
};
>
class B
{
public : B()
{
}
B(const A& ref)
{
printf("destina tion conversion\n");
}
[remainder redactged]

This is homework. vairavans@gmail .com posted the exact same question.

To vairavans and Rahul, please see FAQ 5.2:

https://www.parashift.com/c++-faq-lite/how-to-post.html#faq-5.2

[Rahul]

Re: automatic type conversion

On Dec 12, 12:14 am, red floyd <no.s...@here.d udewrote:

Rahul wrote:
>

class B;

>

class A
{
public : operator B();
};

>

class B
{
public : B()
{
}
B(const A& ref)
{
printf("destina tion conversion\n");
}
[remainder redactged]

>
This is homework. vairav...@gmail .com posted the exact same question.
>
To vairavans and Rahul, please see FAQ 5.2:https://www.parashift.com/c++-faq-li...t.html#faq-5.2

I don't understand what possible home work do u make out of it?
I'm just trying to understand cases where compiler spots an
ambiguity... besides the complexity just increases with operator=,
copy constructor and their combination with source converter or
destination converter or both...
Naturally one would like to clear out the basic simple cases
first...

[red floyd]

Re: automatic type conversion

Rahul wrote:

On Dec 12, 12:14 am, red floyd <no.s...@here.d udewrote:

>Rahul wrote:
>>

>>class B;
>>class A
>>{
>>public : operator B();
>>};
>>class B
>>{
>>public : B()
>> {
>> }
>> B(const A& ref)
>> {
>> printf("destina tion conversion\n");
>> }
>>[remainder redactged]

>This is homework. vairav...@gmail .com posted the exact same question.
>>
>To vairavans and Rahul, please see FAQ 5.2:https://www.parashift.com/c++-faq-li...t.html#faq-5.2

>
I don't understand what possible home work do u make out of it?
I'm just trying to understand cases where compiler spots an
ambiguity... besides the complexity just increases with operator=,
copy constructor and their combination with source converter or
destination converter or both...
Naturally one would like to clear out the basic simple cases
first...

If it's not homework, then why did you and vairavans post identical code
(down to formatting!) with the same question?

[James Kanze]

Re: automatic type conversion

On Dec 11, 7:28 pm, Rahul <sam_...@yahoo. co.inwrote:

I have the following code,

class B;

class A
{
public : operator B();
};

class B
{
public : B()
{
}
B(const A& ref)
{
printf("destina tion conversion\n");
}

B(const B& ref)
{
printf("copy constructor\n") ;
}
};

A::operator B()
{
printf("source conversion\n");
return B();}

int main()
{
A obj;
B obj1 = obj;
return(0);
}

so the output is

destination conversion.

There are actually two ways to convert from obj to obj1
1) destination conversion, constructor function of B
2) source conversion, operator function of A and then copy constructor
of B

The compiler has selected the first option, where as I
expected a ambiguity compile time error, what does the
standard say for this?

That you should see "source conversion" (which is also what I
get from g++ and Sun CC). The requirement in "B obj1 = obj;" is
that obj must be converted into a B (which is then used as an
argument for the copy constructor). A::operator B() is an exact
match, in every detail. B( A const& ) is classified as an exact
match, but does require adding a const, which in this case, the
compiler shoul use as a tie-breaker.

Change the code to:
A const obj ;
B obj1 = obj ;
and you should see "destinatio n conversion", since A::operator
B() can no longer be called on obj. (You'll also have to
provide a user defined default constructor for A.)

Declare A::operator B() const (which it probablyl should be),
and the conversions will be ambiguous (but g++ allows it, using
"destinatio n conversions"---an error in g++).

In practice, of course, it shouldn't matter---you should never
provide round-trip implicit conversions anyway.

--
James Kanze (GABI Software) email:james.kan ze@gmail.com
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