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Copy Constructor?


Copy Constructor?

By : L.s.
Date : October 23 2020, 11:12 PM
hope this fix your issue I assume they are referring to return-value optimization implemented in many compilers where the code:
code :


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Why does the implicit copy constructor calls the base class copy constructor and the defined copy constructor doesn't?

Why does the implicit copy constructor calls the base class copy constructor and the defined copy constructor doesn't?


By : Reza Toorani
Date : March 29 2020, 07:55 AM
I hope this helps . That's just the way the implicit copy constructor is defined (it wouldn't make sense calling the default). As soon as you define any constructor (copy or otherwise) its normal automatic behavior is to call the default parent constructor, so it would be inconsistent to change that for one specific user-defined constructor.
c++ what does Derived Default copy constructor initialize Base copy constructor with?

c++ what does Derived Default copy constructor initialize Base copy constructor with?


By : Abigor
Date : March 29 2020, 07:55 AM
With these it helps First, a cast is something you write in your code to tell the compiler to do a conversion. There are two categories of conversions: implicit and explicit. Implicit conversions will be done when needed, without a cast. Explicit conversions require a cast. What you're talking about here is an implicit conversion, not a cast.
And the answer is that there is an implicit conversion from a reference to a derived type into a reference to a base type. It's that simple: n is a D&, and it can be passed to a function that takes a B& simply by implicitly converting its type.
Conditions under which compiler will not define implicits (constructor, destructor, copy constructor, copy assignment)

Conditions under which compiler will not define implicits (constructor, destructor, copy constructor, copy assignment)


By : Noman
Date : March 29 2020, 07:55 AM
help you fix your problem The Default Constuctor (e.g., X()) will not be implicitly generated if:
you have explicitly declared any constructor at all there is a data member that is not default-constructible (such as a reference, a const object, or a class with no or inaccessible default constructor) (C++11) you have explicitly told the compiler to not generate one using X() = delete;
Why does the C++ copy constructor behave differently when calling its child copy constructor?

Why does the C++ copy constructor behave differently when calling its child copy constructor?


By : Laura Gilbert
Date : March 29 2020, 07:55 AM
will be helpful for those in need When you write a constructor - any constructor - for a derived class you can (and often must) explicitly initialize base class subobjects in that constructor's initializer list. If you fail to do that, then the compiler will implicitly call the default constructors for those base class subobjects, assuming they are available. This rule applies to absolutely all user-defined constructors.
This is exactly what happened in your case. You forgot to initialize base A in B::B(const B&)'s constructor, so the default constructor was used for that base. The fact that B::B(const B&) is a copy constructor makes no difference whatsoever in this case. It works that way, again, consistently for all kinds of user-defined constructors.
Why deleting copy constructor for member variable doesnt prevent defaulting copy constructor

Why deleting copy constructor for member variable doesnt prevent defaulting copy constructor


By : maltz
Date : March 29 2020, 07:55 AM
will help you Declaring a special member as defaulted does not mean that there will be a usable definition. Rather, = default means "fill in the plausible meaning here", and that can also mean that the member ends up being defined as deleted, if that's the only plausible thing (e.g. if class members aren't copyable).
The precise rules are in [class.copy.ctor]:
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