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    <p>First, let's clarify our vocabulary a bit. A default constructor is a constructor which can be called without any arguments. A copy constructor is a constructor which can be called with a single argument of the same type. Given this, a "default copy constructor" would be a constructor with a signature something like: </p> <pre><code>class MyClass { public: static MyClass ourDefaultInstance; // default copy constructor... MyClass( MyClass const&amp; other = ourDefaultInstance ); }; </code></pre> <p>Somehow, I don't think that this is what you meant. I <em>think</em> what you're asking about is an implicitly declared or an implicitly defined copy constructor; a copy constructor whose declaration or definition is provided implicitly by the compiler. The compiler will always provide the declaration unless you provide a declaration of something that can be considered a copy constructor. Providing other constructors will not prevent the compiler from implicitly declaring a copy constructor.</p> <p>This is different from the default constructor&mdash;<em>any</em> user defined constructor will prevent the compiler from implicitly declaring a default constructor. This means that if you have a user defined copy constructor, the compiler will not implicitly declare a default constructor.</p> <p>The second important point is that you do not call constructors. The compiler calls them in certain well defined contexts: variable definition and type conversion, mainly. The compiler can only call constructors that are declared (including those that are implicitly declared). So if you have a user defined constructor (copy or otherwise), and do not define a default constructor, the compiler cannot call the constructor except in contexts where it has arguments to call it with.</p> <p>To summarize what I think your questions are: the compiler will provide an implicit copy constructor even if the class has other user defined constructors, provided none of those constructors can be considered copy constructors. And if you provide a user defined copy constructor, the compiler will <em>not</em> provide an implicitly declared default copy constructor.</p>
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    1. POdefault copy constructor
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    1. USDarrick Herwehe
    1. USJames Kanze
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    1. CO@James Kanze: Very good explaination , by the default copy constructor I meant implicitly defined by the compiler.
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      1. USsourabh912
    2. COAgreed. +1. A "default copy constructor" is a copy constructor that takes a default, nice! That raises an interesting question: How to construct `ourDefaultInstance`. Default constructing it would be UB. I suppose a non-default constructor is needed.
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      1. USDavid Hammen
    3. CO@DavidHammen Yes. I only showed the minimum essential to the point I was making, but anytime you declare a copy constructor, you must declare other constructors as well, or you'll never be able to construct anything to copy. In this case, yes, trying to default construct `ourDefaultInstance` would be undefined behavior.
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