Confused between a copy constructor and a converting constructor

Question

Since I have doubts about this question (for C++03) I am posting it here.I just read about conversion constructors and it states that

"To be a converting constructor, constructor must have single argument and be declared without keyword explicit."

Now my question is whether the copy constructor can be called a conversion constructor provided it is not explicitly declared ? Does it qualify to be one ? I believe it cant be called a conversion constructor because it only accepts the same type parameter ths resulting in no conversion. For Instance

foo a;
foo b;
a = 100; //a Conversion constructor would be called (i.e) foo(int a){...}
a = b ;  //Since both objects are same type and have been initialized the assignment operator will be called (if there is an overloaded version otherwise the default will be called)

Is my understanding correct ?

Answer 1

Quoting the Standard:

[class.conv.ctor]/3

A non-explicit copy-constructor (12.8) is a converting constructor. An implicitly-declared copy constructor is not an explicit constructor; it may be called for implicit type conversions.

So yes, a copy-ctor is a converting ctor.

Also note [conv]/1 which specifies and points out in a remark:

Note: a standard conversion sequence can be empty, i.e., it can consist of no conversions.

and in /3:

An expression e can be implicitly converted to a type T if and only if the declaration T t=e; is well-formed

So the set of implicit conversions contain the empty conversions.

Answer 2

Yes, a copy constructor is what it is -- a copy constructor. Which isn't a conversion constructor that converts from one type to a different type.