Copy Const vs Assignment Operator

[n4nature]

1)
X x;
Y y(x);  //explicit conversion

2)
X x;
Y y=x;//implicit conversion 

How does 2) does an implicit converstion? What are the steps executed by compiler?

Thanks!

[Bazzy]

2) is identical to 1).
In 1) you call the copy constructor with explicit syntax, in 2) you call the same constructor with implicit syntax

[1yR4jE8b]

yeah, both are the same...just different ways to write them.

I try to be explicit as much as possible, I've had some weird things happen with implicit conversions.

[n4nature]

So,

1) if I don't have an explicit Copy Const, could 1) compile (with just an implicit Copy Const)?

2) I think that 2) won't compile if there is a one and only one Copy Const declared, which is explicit. Is that true?

[Bazzy]

If you declare the constructor explicit you can call it only with ()
If you don't you can use =

There's absolutely no difference when using = instead of () for constructors

[n4nature]

Thanks Bazzy!

Do you know why the auto_ptr class have only explicit constructor, and not implicit at all? Wouldn't adding an implicit constructor facilitate initialization as in:

Y y=x

?

[jsmith]

The point is to force the programmer to know what they are doing. Otherwise, think about
it -- EVERY single raw pointer you hold in your program could be implicitly convertible to
a std::auto_ptr.

[n4nature]

Thank you!