Re: Compiler chooses conv ctor - why?

Hi Alex,
Could you please let me know why you think that this is a compiler bug? I
think that this is just a way of compiler implementing conversion. In VC++,
by default "B b=a" will call B's constructor "B(const A& a);" to perform
the conversion; you can use "explicit" keyword before "B(const A& a);" to
avoid such automatic constructor conversion, and then the conversion would
happen on "A::operator const B() const".

Hi Cezary,
Regarding your original question, why compiler choosing conv ctor, I think
that it is related to some implicit conversion rule for compiler. When
multiple conversion methods are detected, one must be chose as with the
first priority. Unfortunately the detail of this is undocumented. Based on
my test, I think that VC++ compiler by default uses conversion constructor
prior to operator conversion. If you comment the constructor "B(const A&
a);", you will find that all conversions use your operator conversion
"A::operator const B() const" instead.

We can find some clues from MSDB documents:
Drawbacks of Conversion Constructors
http://msdn2.microsoft.com/en-us/library/1zd41sht.aspx

As the article mentioned, if it is essential to retain full control, do not
declare any constructors that take a single argument; instead, define
"helper" functions to perform conversions. So I would recommend that you
refer to this article as the best practice when you program in VC++.

If you have any other questions or concerns, please feel free to let me
know.

Best regards,
Charles Wang
Microsoft Online Community Support
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