Is there a default explicit conversion function for user-defined class (c++)?

Question

Can I convert compatible non-pointer objects without actually defining a cast operator? So if B inherits from A:

A a;
B b;

Will (A)b compile if I donot provide my own cast operator?

Edit: It does seem to compile. Is it because there is a default conversion cast operator or the compiler recognizes the compatibility and uses the default copying constructor or assignment operator in assignments for example?

Answer 1

Yes, it does compile (http://ideone.com/CXbeh0)

By default A is copyable. The copy constructor has signature A::A(A const &). Since instance of B is implicitly convertible to A const &, the cast will be resolved using A's copy constructor.

The conversion to reference applies to the (also implicitly generated) copy assignment operator too, so a = b again compiles using A &A::operator=(A const &).

Keep in mind, that the new object is of type A, not B, so it does not contain any additional information the original instance of B did. That is rarely what you want.

On a side-note, it is recommended in C++ to forget that C-style cast exists and use the more specific cast types of C++:

• The function-style cast for explicit request to create a temporary of target type using matching constructor; also allows using multi-parameter constructors.
• The static_cast, for compatible types only.
• The dynamic_cast for upcasting pointers/references with run-time check.
• const_cast to only handle const.
• And reinterpret_cast if you really need to play pointer tricks, but beware of aliasing rules.

Answer 2

By no user-defined constructors or operators are defined, then:

(A)b

will be implemented as if:

A(b) //Calls A copy-constructor: A::A(const A&)

That is, it will create a temporary of A copy-constructing from the argument.

This is called object splicing and it is usually a bad idea, because you are copying some fields of B, but not others. YMMV.