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POStrange copy constructor
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  1. POStrange copy constructor
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    copied!<p>With the following C++ program:</p> <pre><code>#include &lt;memory&gt; #include &lt;iostream&gt; using namespace std; struct my_class{ int value; my_class(int id): value(id){ cout&lt;&lt;"constructing "&lt;&lt;id&lt;&lt;endl; cout&lt;&lt;"address is "&lt;&lt;static_cast&lt;const void *&gt;(this)&lt;&lt;endl; } my_class(const my_class &amp; a){ cout&lt;&lt;"construct copying "&lt;&lt;a.value&lt;&lt;endl; cout&lt;&lt;static_cast&lt;const void *&gt;(this)&lt;&lt;"&lt;-"&lt;&lt;static_cast&lt;const void *&gt;(&amp;a)&lt;&lt;endl; } my_class operator=(const my_class &amp; a){ cout&lt;&lt;"assignment copying "&lt;&lt;a.value&lt;&lt;endl; this-&gt;value = a.value; cout&lt;&lt;static_cast&lt;const void *&gt;(this)&lt;&lt;"&lt;-"&lt;&lt;static_cast&lt;const void *&gt;(&amp;a)&lt;&lt;endl; return *this; } ~my_class(){ cout&lt;&lt;"deleting "&lt;&lt;this-&gt;value&lt;&lt;endl; cout&lt;&lt;"address is "&lt;&lt;static_cast&lt;const void *&gt;(this)&lt;&lt;endl; } }; my_class f(){ cout&lt;&lt;"==in f=="&lt;&lt;endl; my_class temp(2); cout&lt;&lt;"==out f=="&lt;&lt;endl; return temp; } int main(){ cout&lt;&lt;"==in main=="&lt;&lt;endl; my_class a(1); a = f(); a.value++; cout&lt;&lt;"==out main=="&lt;&lt;endl; return 0; } </code></pre> <p>I got the following result:</p> <pre><code>==== ==in main== constructing 1 address is 0x28ff04 ==in f== constructing 2 address is 0x28ff0c ==out f== assignment copying 2 0x28ff04&lt;-0x28ff0c construct copying 2 0x28ff08&lt;-0x28ff04 deleting 2686868 address is 0x28ff08 deleting 2 address is 0x28ff0c ==out main== deleting 3 address is 0x28ff04 === </code></pre> <p>Can anyone explain to me what happens with the object at the address "0x28ff08" and the related copy constructing from object at the address "0x28ff04"? I really don't understand why the copy constructor is called here.</p> <hr> <p>I don't known if I get this correct, thus I want to further explain it in great detail. Anyone finds my mistakes please point them out.</p> <p>First, an image illustrates the details of the execution flow: <img src="https://i.stack.imgur.com/8LZPP.png" alt="the execution flow"></p> <p>(1). Create an object <code>a</code> with value 1;</p> <p>(2). Call function <code>f()</code>. Create an object <code>temp</code>, and the compiler finds that the object will be returned, so it is created directly in the caller's stack;</p> <p>(3). Assign the returned object of <code>f()</code> (i.e., <code>temp</code>) to object <code>a</code> by calling <code>operator=()</code> of <code>a</code>;</p> <p>(4). Object <code>a</code> is passed into <code>operator=()</code> as the parameter (rvalue) using the same variable name <code>a</code>.</p> <p>(5). The method <code>operator=()</code> is called on <code>main::a</code> (lvalue, with the abuse of notation), thus <code>this</code> in the function points to <code>main::a</code>, [!!this is the part that confused me];</p> <p>(6). <code>operator=()</code> changes the value of <code>main::a</code> to <code>a</code>'s value (i.e., from 1 to 2);</p> <p>(7). The compiler finds that the return type is not a reference, and <code>*this</code> already exists in <code>main()</code>, so it has to copy <code>*this</code> by calling the copy constructor. However, the copy constructor does not initialize the object, so an uninitialized object is created.</p> <p>(8). [!!not quite sure about this part] The lvalue and the resulting object is the same object, thus no object is really returned due to optimization.</p> <p>(9). The copied object is destroyed, according to @Mike Seymour, this object is created because the compiler cannot omit it because both the constructor and destructor actually do something (e.g. output the value and address).</p> <p>(10). When exiting <code>operator=()</code>, object <code>a</code> is destroyed.</p> <p>(11). When exiting <code>main()</code>, object <code>main::a</code> is finally destroyed.</p> <p>The above explains the output, however, my current understanding may not be correct. Please help me comprehend this if I am wrong. Thanks a lot.</p>
 

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