question about "operator ="

[tomy]

Hi All:
I just wonder what happened when i use operator "=" to construct an
obj, as following:

class Obj
{
public:
Obj();
Obj& operator = (const Obj &);
}
Obj a;
Obj b = a;

The Obj b will be constructed by "operator =" directly,
or first by "Obj()" , then turn to "operator =", two steps.

I've write some "cout" in both of the functions, with gcc, no print any
more;
But finally the Obj b contain the same value with Obj a, obviously;

So what happend?

Thanks

 

[Victor Bazarov]

Hi All:
I just wonder what happened when i use operator "=" to construct an
obj, as following:

class Obj
{
public:
Obj();
Obj& operator = (const Obj &);
}
Obj a;
Obj b = a;

This syntax is called copy-initialisation, it has nothing to do
with assignment except using the same token, '='.

The Obj b will be constructed by "operator =" directly,

No, it will be constructed by copy-constructor directly.

or first by "Obj()" , then turn to "operator =", two steps.
No.

I've write some "cout" in both of the functions, with gcc, no print
any more;
But finally the Obj b contain the same value with Obj a, obviously;

So what happend?

Copy-initialisation. Isn't that what your favourite C++ book says?

V

 

[tomy]

This syntax is called copy-initialisation, it has nothing to do
with assignment except using the same token, '='.


No, it will be constructed by copy-constructor directly.


Copy-initialisation. Isn't that what your favourite C++ book says? Wow~~~, That's it.

V