Question about conversion in expressions

[Jeroen]

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

[terminator]

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.
};