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copied!<p><strong>Shallow copy:</strong></p> <p>Some members of the copy may reference the same objects as the original:</p> <pre><code>class X { private: int i; int *pi; public: X() : pi(new int) { } X(const X& copy) // <-- copy ctor : i(copy.i), pi(copy.pi) { } }; </code></pre> <p>Here, the <code>pi</code> member of the original and copied <code>X</code> object will both point to the same <code>int</code>.</p> <hr> <p><strong>Deep copy:</strong></p> <p>All members of the original are cloned (recursively, if necessary). There are no shared objects:</p> <pre><code>class X { private: int i; int *pi; public: X() : pi(new int) { } X(const X& copy) // <-- copy ctor : i(copy.i), pi(new int(*copy.pi)) // <-- note this line in particular! { } }; </code></pre> <p>Here, the <code>pi</code> member of the original and copied <code>X</code> object will point to different <code>int</code> objects, but both of these have the same value.</p> <hr> <p><strike><sub>The default copy constructor (which is automatically provided if you don't provide one yourself) creates only shallow copies.</sub></strike></p> <blockquote> <p><strong>Correction:</strong> Several comments below have correctly pointed out that it is wrong to say that the default copy constructor <em>always</em> performs a shallow copy (or a deep copy, for that matter). Whether a type's copy constructor creates a shallow copy, or deep copy, <strong>or something in-between the two</strong>, depends on the combination of each member's copy behaviour; a member's type's copy constructor can be made to do whatever it wants, after all.</p> </blockquote> <p>Here's what section 12.8, paragraph 8 of the 1998 C++ standard says about the above code examples:</p> <blockquote> <p><em>The implicitly defined copy constructor for class <code>X</code> performs a memberwise copy of its subobjects. [...] Each subobject is copied in the manner appropriate to its type: [...] [I]f the subobject is of scalar type, the builtin assignment operator is used.</em></p> </blockquote>

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