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Question about conversion in expressions

Hi guys,

Just another question. Suppose I have 2 classes (incomplete code):

class A {
A(const B& b);
A& operator = (const A& a);
};

class B {
B(const A& a);
friend B operator + (const B& b1, const B& b2);
};

If I have code like:

{
A x, y, z;

x = y + z;
}

There is no + operator for A. Is it possible that the ctor of B makes
temporary objects of 'y' and 'z', then the + operator of B is called to
add x and y, and the result is converted back to class A with its ctor
and then assigned to x? Or am I talking complete nonsense here?

Thanx for any pointers,

Jeroen
Feb 28 '07 #1
10 1439
On Feb 28, 4:50 pm, Jeroen <no_m...@thanx.comwrote:
Hi guys,

Just another question. Suppose I have 2 classes (incomplete code):

class A {
A(const B& b);
A& operator = (const A& a);

};

class B {
B(const A& a);
friend B operator + (const B& b1, const B& b2);

};

If I have code like:

{
A x, y, z;

x = y + z;

}

There is no + operator for A. Is it possible that the ctor of B makes
temporary objects of 'y' and 'z', then the + operator of B is called to
add x and y, and the result is converted back to class A with its ctor
and then assigned to x? Or am I talking complete nonsense here?

Thanx for any pointers,

Jeroen
The compiler can not deduce how it should add z and y ;there maybe
other classes similar to B too.I suggest either of the following
lines:

x=B(y)+z;
x=y+B(z);

in either of these cases in contrast to wht you did the compiler is
able to pick 'B operator+(const B&,const B&)'.

Feb 28 '07 #2
terminator schreef:
On Feb 28, 4:50 pm, Jeroen <no_m...@thanx.comwrote:
>Hi guys,

Just another question. Suppose I have 2 classes (incomplete code):

class A {
A(const B& b);
A& operator = (const A& a);

};

class B {
B(const A& a);
friend B operator + (const B& b1, const B& b2);

};

If I have code like:

{
A x, y, z;

x = y + z;

}

There is no + operator for A. Is it possible that the ctor of B makes
temporary objects of 'y' and 'z', then the + operator of B is called to
add x and y, and the result is converted back to class A with its ctor
and then assigned to x? Or am I talking complete nonsense here?

Thanx for any pointers,

Jeroen

The compiler can not deduce how it should add z and y ;there maybe
other classes similar to B too.I suggest either of the following
lines:

x=B(y)+z;
x=y+B(z);

in either of these cases in contrast to wht you did the compiler is
able to pick 'B operator+(const B&,const B&)'.
OK, so at least one variable should match the available arithmetic
operator to ensure that it works. Thanks!
Feb 28 '07 #3

"Jeroen" <no*****@thanx.comwrote in message news:45e58df7@cs1...
terminator schreef:
>On Feb 28, 4:50 pm, Jeroen <no_m...@thanx.comwrote:
>>Hi guys,

Just another question. Suppose I have 2 classes (incomplete code):

class A {
A(const B& b);
A& operator = (const A& a);

};

class B {
B(const A& a);
friend B operator + (const B& b1, const B& b2);

};

If I have code like:

{
A x, y, z;

x = y + z;

}

There is no + operator for A. Is it possible that the ctor of B makes
temporary objects of 'y' and 'z', then the + operator of B is called to
add x and y, and the result is converted back to class A with its ctor
and then assigned to x? Or am I talking complete nonsense here?

Thanx for any pointers,

Jeroen

The compiler can not deduce how it should add z and y ;there maybe
other classes similar to B too.I suggest either of the following
lines:

x=B(y)+z;
x=y+B(z);

in either of these cases in contrast to wht you did the compiler is
able to pick 'B operator+(const B&,const B&)'.

OK, so at least one variable should match the available arithmetic
operator to ensure that it works. Thanks!
But the question that you should rather ask yourself is should I really do
this? You're walking towards implicit conversion, which is a dangerous
thing, and also towards a rather questionable design. If A and B are not
related, which is what is indicated by your design, then why would you want
to convert A to B and probably back. It's even more questionable why you
want the compiler to do this silently. If A & B are not related then they
also should not share implementations, like the + op for example. I'd
strongly suggest that you reconsider your design because IMHO this looks
very much like the beginning of a serious design flaw, which could provide
you the arguable pleasure of endless debugging hours.

Cheers
Chris
Feb 28 '07 #4
Chris Theis wrote:
"Jeroen" <no*****@thanx.comwrote in message news:45e58df7@cs1...
>terminator schreef:
>>On Feb 28, 4:50 pm, Jeroen <no_m...@thanx.comwrote:
Hi guys,

Just another question. Suppose I have 2 classes (incomplete code):

class A {
A(const B& b);
A& operator = (const A& a);

};

class B {
B(const A& a);
friend B operator + (const B& b1, const B& b2);

};

If I have code like:

{
A x, y, z;

x = y + z;

}

There is no + operator for A. Is it possible that the ctor of B makes
temporary objects of 'y' and 'z', then the + operator of B is called to
add x and y, and the result is converted back to class A with its ctor
and then assigned to x? Or am I talking complete nonsense here?

Thanx for any pointers,

Jeroen
The compiler can not deduce how it should add z and y ;there maybe
other classes similar to B too.I suggest either of the following
lines:

x=B(y)+z;
x=y+B(z);

in either of these cases in contrast to wht you did the compiler is
able to pick 'B operator+(const B&,const B&)'.
OK, so at least one variable should match the available arithmetic
operator to ensure that it works. Thanks!

But the question that you should rather ask yourself is should I really do
this? You're walking towards implicit conversion, which is a dangerous
thing,
Implicit conversion isn't inherently dangerous. It occurs, for example,
whenever you pass an object of a derived type to a function that takes a
reference to one of its base types. And implicit conversions among value
types isn't necessarily dangerous, either.

The real issue here is the interconvertibility of the two types, which
often leads to mysterious-looking ambiguities.
and also towards a rather questionable design. If A and B are not
related, which is what is indicated by your design,
Seems to me that the sketch we've seen indicates that A and B are
closely related.

--

-- Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com)
Author of "The Standard C++ Library Extensions: a Tutorial and
Reference." (www.petebecker.com/tr1book)
Feb 28 '07 #5
Pete Becker schreef:
Chris Theis wrote:
>"Jeroen" <no*****@thanx.comwrote in message news:45e58df7@cs1...
>>terminator schreef:
On Feb 28, 4:50 pm, Jeroen <no_m...@thanx.comwrote:
Hi guys,
>
Just another question. Suppose I have 2 classes (incomplete code):
>
class A {
A(const B& b);
A& operator = (const A& a);
>
};
>
class B {
B(const A& a);
friend B operator + (const B& b1, const B& b2);
>
};
>
If I have code like:
>
{
A x, y, z;
>
x = y + z;
>
}
>
There is no + operator for A. Is it possible that the ctor of B makes
temporary objects of 'y' and 'z', then the + operator of B is
called to
add x and y, and the result is converted back to class A with its ctor
and then assigned to x? Or am I talking complete nonsense here?
>
Thanx for any pointers,
>
Jeroen
The compiler can not deduce how it should add z and y ;there maybe
other classes similar to B too.I suggest either of the following
lines:

x=B(y)+z;
x=y+B(z);

in either of these cases in contrast to wht you did the compiler is
able to pick 'B operator+(const B&,const B&)'.

OK, so at least one variable should match the available arithmetic
operator to ensure that it works. Thanks!

But the question that you should rather ask yourself is should I
really do this? You're walking towards implicit conversion, which is a
dangerous thing,

Implicit conversion isn't inherently dangerous. It occurs, for example,
whenever you pass an object of a derived type to a function that takes a
reference to one of its base types. And implicit conversions among value
types isn't necessarily dangerous, either.

The real issue here is the interconvertibility of the two types, which
often leads to mysterious-looking ambiguities.
>and also towards a rather questionable design. If A and B are not
related, which is what is indicated by your design,

Seems to me that the sketch we've seen indicates that A and B are
closely related.
Indeed they are closely related. B contains a pointer to an 'A' object,
and some additional information. I'm playing around with possible
implementations which may solve several problems of my project. I'm not
yet up to actually implement things. I'll start with that once I know
that I have an elegant architecture which will work without problems.
See my respond in the previous thread 'Question about evaluation order'
for some info of what I'm doing.

Thanx,

Jeroen
Feb 28 '07 #6
On Feb 28, 5:02 pm, "terminator" <farid.mehr...@gmail.comwrote:
On Feb 28, 4:50 pm, Jeroen <no_m...@thanx.comwrote:


Hi guys,
Just another question. Suppose I have 2 classes (incomplete code):
class A {
A(const B& b);
A& operator = (const A& a);
};
class B {
B(const A& a);
friend B operator + (const B& b1, const B& b2);
};
If I have code like:
{
A x, y, z;
x = y + z;
}
There is no + operator for A. Is it possible that the ctor of B makes
temporary objects of 'y' and 'z', then the + operator of B is called to
add x and y, and the result is converted back to class A with its ctor
and then assigned to x? Or am I talking complete nonsense here?
Thanx for any pointers,
Jeroen

The compiler can not deduce how it should add z and y ;there maybe
other classes similar to B too.I suggest either of the following
lines:

x=B(y)+z;
x=y+B(z);

in either of these cases in contrast to wht you did the compiler is
able to pick 'B operator+(const B&,const B&)'.- Hide quoted text -

- Show quoted text -
Your code works unless another class that similar to B defines a
global binary operator+ and a constructor from an A is introduced.

Feb 28 '07 #7
On Feb 28, 8:25 pm, "terminator" <farid.mehr...@gmail.comwrote:
On Feb 28, 5:02 pm, "terminator" <farid.mehr...@gmail.comwrote:


On Feb 28, 4:50 pm, Jeroen <no_m...@thanx.comwrote:
Hi guys,
Just another question. Suppose I have 2 classes (incomplete code):
class A {
A(const B& b);
A& operator = (const A& a);
};
class B {
B(const A& a);
friend B operator + (const B& b1, const B& b2);
};
If I have code like:
{
A x, y, z;
x = y + z;
}
There is no + operator for A. Is it possible that the ctor of B makes
temporary objects of 'y' and 'z', then the + operator of B is called to
add x and y, and the result is converted back to class A with its ctor
and then assigned to x? Or am I talking complete nonsense here?
Thanx for any pointers,
Jeroen
The compiler can not deduce how it should add z and y ;there maybe
other classes similar to B too.I suggest either of the following
lines:
x=B(y)+z;
x=y+B(z);
in either of these cases in contrast to wht you did the compiler is
able to pick 'B operator+(const B&,const B&)'.- Hide quoted text -
- Show quoted text -

Your code works unless another class that similar to B defines a
global binary operator+ and a constructor from an A is introduced.- Hide quoted text -

- Show quoted text -
class B;
class A{
public:
A(){cout<<"A()\n";};
A(const B&){cout<<"A(B)\n";};
};

class B{
public:
B(){cout<<"B()\n";};
B(const A&){cout<<"B(A)\n";};
friend B operator+(const B&,const B&);
};

B operator+(const B&,const B&){cout<<"B+B\n";return B();};

//if you uncomment the following ,ambiguity error occurs.
/*class C{
public:
C(){cout<<"C()\n";};
C(const A&){cout<<"C(A)\n";};
friend C operator+(const C&,const C&);
};

C operator+(const C&,const C&){cout<<"C+C\n";return C();};*/

int _tmain(int argc, _TCHAR* argv[])
{
A x,y,z;
x=y+z;//ok:x=A(B(y)+B(z));
return 0;
}

Feb 28 '07 #8

"Pete Becker" <pe**@versatilecoding.comwrote in message
news:g-******************************@giganews.com...
Implicit conversion isn't inherently dangerous. It occurs, for example,
whenever you pass an object of a derived type to a function that takes a
reference to one of its base types. And implicit conversions among value
types isn't necessarily dangerous, either.
I absolutely agree that it's not inherently or necessarily dangerous. But
IMO one should be careful as it might open pitfalls that are very often not
obvious and hard to spot. Still, it certainly has its place and uses. There
is no doubt about that.

Cheers
Chris
Feb 28 '07 #9
terminator schreef:
On Feb 28, 8:25 pm, "terminator" <farid.mehr...@gmail.comwrote:
>On Feb 28, 5:02 pm, "terminator" <farid.mehr...@gmail.comwrote:


>>On Feb 28, 4:50 pm, Jeroen <no_m...@thanx.comwrote:
Hi guys,
Just another question. Suppose I have 2 classes (incomplete code):
class A {
A(const B& b);
A& operator = (const A& a);
};
class B {
B(const A& a);
friend B operator + (const B& b1, const B& b2);
};
If I have code like:
{
A x, y, z;
x = y + z;
}
There is no + operator for A. Is it possible that the ctor of B makes
temporary objects of 'y' and 'z', then the + operator of B is called to
add x and y, and the result is converted back to class A with its ctor
and then assigned to x? Or am I talking complete nonsense here?
Thanx for any pointers,
Jeroen
The compiler can not deduce how it should add z and y ;there maybe
other classes similar to B too.I suggest either of the following
lines:
x=B(y)+z;
x=y+B(z);
in either of these cases in contrast to wht you did the compiler is
able to pick 'B operator+(const B&,const B&)'.- Hide quoted text -
- Show quoted text -
Your code works unless another class that similar to B defines a
global binary operator+ and a constructor from an A is introduced.- Hide quoted text -

- Show quoted text -

class B;
class A{
public:
A(){cout<<"A()\n";};
A(const B&){cout<<"A(B)\n";};
};

class B{
public:
B(){cout<<"B()\n";};
B(const A&){cout<<"B(A)\n";};
friend B operator+(const B&,const B&);
};

B operator+(const B&,const B&){cout<<"B+B\n";return B();};

//if you uncomment the following ,ambiguity error occurs.
/*class C{
public:
C(){cout<<"C()\n";};
C(const A&){cout<<"C(A)\n";};
friend C operator+(const C&,const C&);
};

C operator+(const C&,const C&){cout<<"C+C\n";return C();};*/

int _tmain(int argc, _TCHAR* argv[])
{
A x,y,z;
x=y+z;//ok:x=A(B(y)+B(z));
return 0;
}
OK, thanks for the example. If I get this right, then C++ will try to do
any conversion in order to find a corresponding match with the required
operator, but it is a dangerous technique... I will avoid it :-)

Jeroen
Mar 1 '07 #10
On Mar 1, 12:03 pm, Jeroen <no_m...@thanx.comwrote:
terminator schreef:


On Feb 28, 8:25 pm, "terminator" <farid.mehr...@gmail.comwrote:
On Feb 28, 5:02 pm, "terminator" <farid.mehr...@gmail.comwrote:
>On Feb 28, 4:50 pm, Jeroen <no_m...@thanx.comwrote:
Hi guys,
Just another question. Suppose I have 2 classes (incomplete code):
class A {
A(const B& b);
A& operator = (const A& a);
};
class B {
B(const A& a);
friend B operator + (const B& b1, const B& b2);
};
If I have code like:
{
A x, y, z;
x = y + z;
}
There is no + operator for A. Is it possible that the ctor of B makes
temporary objects of 'y' and 'z', then the + operator of B is called to
add x and y, and the result is converted back to class A with its ctor
and then assigned to x? Or am I talking complete nonsense here?
Thanx for any pointers,
Jeroen
The compiler can not deduce how it should add z and y ;there maybe
other classes similar to B too.I suggest either of the following
lines:
x=B(y)+z;
x=y+B(z);
in either of these cases in contrast to wht you did the compiler is
able to pick 'B operator+(const B&,const B&)'.- Hide quoted text -
- Show quoted text -
Your code works unless another class that similar to B defines a
global binary operator+ and a constructor from an A is introduced.- Hide quoted text -
- Show quoted text -
class B;
class A{
public:
A(){cout<<"A()\n";};
A(const B&){cout<<"A(B)\n";};
};
class B{
public:
B(){cout<<"B()\n";};
B(const A&){cout<<"B(A)\n";};
friend B operator+(const B&,const B&);
};
B operator+(const B&,const B&){cout<<"B+B\n";return B();};
//if you uncomment the following ,ambiguity error occurs.
/*class C{
public:
C(){cout<<"C()\n";};
C(const A&){cout<<"C(A)\n";};
friend C operator+(const C&,const C&);
};
C operator+(const C&,const C&){cout<<"C+C\n";return C();};*/
int _tmain(int argc, _TCHAR* argv[])
{
A x,y,z;
x=y+z;//ok:x=A(B(y)+B(z));
return 0;
}

OK, thanks for the example. If I get this right, then C++ will try to do
any conversion in order to find a corresponding match with the required
operator, but it is a dangerous technique... I will avoid it :-)

Jeroen- Hide quoted text -

- Show quoted text -
if you define 'B' like this:

class B{
public:
B();
B(const A&);
B operator+(const B&); //binary operator as member function
};

then you have to write the code this way:

{
A x,y,z;
x=B(x)+z;//OK:x=A(B(x).B::operator+(B(z)));
x=y+z;//ERROR:no + for A.
};
Mar 3 '07 #11

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